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Numbered Examples:
Chapters One to Eight
Number and Topic
Number and Topic
Chapter One
4.7–1
1.1–1
1.6–1
Volume of a circular cylinder
Piston motion
Chapter Two
2.3–1
2.3–2
2.3–3
2.3–4
2.3–5
2.4–1
2.4–2
2.4–3
2.4–4
2.5–1
2.6–1
2.7–1
Vectors and displacement
Aortic pressure model
Transportation route analysis
Current and power dissipation in
resistors
A batch distillation process
Miles traveled
Height versus velocity
Manufacturing cost analysis
Product cost analysis
Earthquake-resistant building design
An environmental database
A student database
Chapter Three
3.2–1
Optimization of an irrigation channel
Chapter Four
4.3–1
4.5–1
4.5–2
4.5–3
4.5–4
4.6–1
4.6–2
4.6–3
Height and speed of a projectile
Series calculation with a for loop
Plotting with a for loop
Data sorting
Flight of an instrumented rocket
Series calculation with a while loop
Growth of a bank account
Time to reach a speci ed height
4.9–1
4.9–2
Using the switch structure for calendar
calculations
A college enrollment model: Part I
A college enrollment model: Part II
Chapter Five
5.2–1
Plotting orbits
Chapter Six
6.1–1
6.1–2
6.2–1
6.2–2
6.2–3
6.2–4
Temperature dynamics
Hydraulic resistance
Estimation of traf c ow
Modeling bacteria growth
Breaking strength and alloy
composition
Response of a biomedical instrument
Chapter Seven
7.1–1
7.2–1
7.2–2
7.3–1
Breaking strength of thread
Mean and standard deviation of heights
Estimation of height distribution
Statistical analysis and manufacturing
tolerances
Chapter Eight
8.1–1
8.2–1
8.2–2
8.2–3
8.2–4
The matrix inverse method
Left division method with three
unknowns
Calculations of cable tension
An electric resistance network
Ethanol production
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Numbered Examples:
Chapters Eight to Eleven
Number and Topic
Number and Topic
8.3–1
Chapter Ten
8.3–2
8.3–3
8.3–4
8.3–5
8.4–1
8.4–2
An underdetermined set with three
equations and three unknowns
A statically indeterminate problem
Three equations in three unknowns,
continued
Production planning
Traf c engineering
The least-squares method
An overdetermined set
Chapter Nine
9.1–1
9.1–2
9.1–3
9.3–1
9.3–2
9.4–1
9.5–1
Velocity from an accelerometer
Evaluation of Fresnel’s cosine integral
Double integral over a nonrectangular
region
Response of an RC circuit
Liquid height in a spherical tank
A nonlinear pendulum model
Trapezoidal pro le for a dc motor
10.2–1
10.2–2
10.2–3
10.3–1
10.4–1
10.4–2
10.5–1
10.6–1
#
Simulink solution of y = 10 sin t
Exporting to the MATLAB workspace
#
Simulink model for y = - 10y + f (t)
Simulink model of a two-mass
suspension system
Simulink model of a rocket-propelled
sled
Model of a relay-controlled motor
Response with a dead zone
Model of a nonlinear pendulum
Chapter Eleven
11.3–1
11.3–2
11.5–1
Intersection of two circles
Positioning a robot arm
Topping the Green Monster
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Introduction to MATLAB®
for Engineers
William J. Palm III
University of Rhode Island
TM
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TM
INTRODUCTION TO MATLAB® FOR ENGINEERS, THIRD EDITION
Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas,
New York, NY 10020. Copyright © 2011 by The McGraw-Hill Companies, Inc. All rights reserved. Previous
editions © 2005 and 2001. No part of this publication may be reproduced or distributed in any form or by any
means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill
Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or
broadcast for distance learning.
Some ancillaries, including electronic and print components, may not be available to customers outside the
United States.
This book is printed on acid-free paper containing 10% postconsumer waste.
1 2 3 4 5 6 7 8 9 0 DOC/DOC 1 0 9 8 7 6 5 4 3 2 1 0
ISBN 978-0-07-353487-9
MHID 0-07-353487-0
Vice President & Editor-in-Chief: Martin Lange
Vice President, EDP: Kimberly Meriwether David
Global Publisher: Raghu Srinivasan
Sponsoring Editor: Bill Stenquist
Marketing Manager: Curt Reynolds
Development Editor: Lora Neyens
Senior Project Manager: Joyce Watters
Design Coordinator: Margarite Reynolds
Cover Designer: Rick D. Noel
Photo Research: John Leland
Cover Image: © Ingram Publishing/AGE Fotostock
Production Supervisor: Nicole Baumgartner
Media Project Manager: Joyce Watters
Compositor: MPS Limited, A Macmillan Company
Typeface: 10/12 Times Roman
Printer: RRDonnelly
All credits appearing on page or at the end of the book are considered to be an extension of the copyright page.
Library of Congress Cataloging-in-Publication Data
Palm, William J. (William John), 1944–
Introduction to MATLAB for engineers / William J. Palm III.—3rd ed.
p. cm.
Includes bibliographical references and index.
ISBN 978-0-07-353487-9
1. MATLAB. 2. Numerical analysis—Data processing. I. Title.
QA297.P33 2011
518.0285—dc22
2009051876
www.mhhe.com
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To my sisters, Linda and Chris, and to my parents, Lillian and William
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ABOUT THE AUTHOR
is Professor of Mechanical Engineering at the University of
Rhode Island. In 1966 he received a B.S. from Loyola College in Baltimore, and
in 1971 a Ph.D. in Mechanical Engineering and Astronautical Sciences from
Northwestern University in Evanston, Illinois.
During his 38 years as a faculty member, he has taught 19 courses. One of
these is a freshman MATLAB course, which he helped develop. He has authored
eight textbooks dealing with modeling and simulation, system dynamics, control
systems, and MATLAB. These include System Dynamics, 2nd Edition (McGrawHill, 2010). He wrote a chapter on control systems in the Mechanical Engineers’
Handbook (M. Kutz, ed., Wiley, 1999), and was a special contributor to the fth
editions of Statics and Dynamics, both by J. L. Meriam and L. G. Kraige (Wiley,
2002).
Professor Palm’s research and industrial experience are in control systems,
robotics, vibrations, and system modeling. He was the Director of the Robotics
Research Center at the University of Rhode Island from 1985 to 1993, and is the
coholder of a patent for a robot hand. He served as Acting Department Chair
from 2002 to 2003. His industrial experience is in automated manufacturing;
modeling and simulation of naval systems, including underwater vehicles and
tracking systems; and design of control systems for underwater-vehicle enginetest facilities.
William J. Palm III
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CONTENTS
Preface ix
CHAPTER
CHAPTER
Programming with MATLAB 147
1
An Overview of MATLAB® 3
1.1 MATLAB Interactive Sessions 4
1.2 Menus and the Toolbar 16
1.3 Arrays, Files, and Plots 18
1.4 Script Files and the Editor/Debugger 27
1.5 The MATLAB Help System 33
1.6 Problem-Solving Methodologies 38
1.7 Summary 46
Problems 47
CHAPTER
2
Numeric, Cell, and Structure Arrays
53
2.1
One- and Two-Dimensional Numeric
Arrays 54
2.2 Multidimensional Numeric Arrays 63
2.3 Element-by-Element Operations 64
2.4 Matrix Operations 73
2.5 Polynomial Operations Using Arrays 85
2.6 Cell Arrays 90
2.7 Structure Arrays 92
2.8 Summary 96
Problems 97
CHAPTER
4
4.1
4.2
Program Design and Development 148
Relational Operators and Logical
Variables 155
4.3 Logical Operators and Functions 157
4.4 Conditional Statements 164
4.5 for Loops 171
4.6 while Loops 183
4.7 The switch Structure 188
4.8 Debugging MATLAB Programs 190
4.9 Applications to Simulation 193
4.10 Summary 199
Problems 200
CHAPTER
5
Advanced Plotting 219
5.1
5.2
xy Plotting Functions 219
Additional Commands and
Plot Types 226
5.3 Interactive Plotting in MATLAB
5.4 Three-Dimensional Plots 246
5.5 Summary 251
Problems 251
3
6
Functions and Files 113
CHAPTER
3.1 Elementary Mathematical Functions 113
3.2 User-De ned Functions 119
3.3 Additional Function Topics 130
3.4 Working with Data Files 138
3.5 Summary 140
Problems 140
6.1 Function Discovery 263
6.2 Regression 271
6.3 The Basic Fitting Interface 282
6.4 Summary 285
Problems 286
Model Building and Regression
241
263
vii
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Contents
viii
CHAPTER
7
10.7
10.8
10.9
Statistics, Probability, and
Interpolation 295
7.1 Statistics and Histograms 296
7.2 The Normal Distribution 301
7.3 Random Number Generation 307
7.4 Interpolation 313
7.5 Summary 322
Problems 324
CHAPTER
8
Linear Algebraic Equations
331
8.1 Matrix Methods for Linear Equations 332
8.2 The Left Division Method 335
8.3 Underdetermined Systems 341
8.4 Overdetermined Systems 350
8.5 A General Solution Program 354
8.6 Summary 356
Problems 357
CHAPTER
9
Numerical Methods for Calculus and
Differential Equations 369
9.1 Numerical Integration 370
9.2 Numerical Differentiation 377
9.3 First-Order Differential Equations 382
9.4 Higher-Order Differential Equations 389
9.5 Special Methods for Linear Equations 395
9.6 Summary 408
Problems 410
CHAPTER
Simulink
10.1
10.2
10.3
10.4
10.5
10.6
10
419
Simulation Diagrams 420
Introduction to Simulink 421
Linear State-Variable Models 427
Piecewise-Linear Models 430
Transfer-Function Models 437
Nonlinear State-Variable Models 441
Subsystems 443
Dead Time in Models 448
Simulation of a Nonlinear Vehicle
Suspension Model 451
10.10 Summary 455
Problems 456
CHAPTER
MuPAD
11
465
11.1
11.2
11.3
Introduction to MuPAD 466
Symbolic Expressions and Algebra 472
Algebraic and Transcendental
Equations 479
11.4 Linear Algebra 489
11.5 Calculus 493
11.6 Ordinary Differential Equations 501
11.7 Laplace Transforms 506
11.8 Special Functions 512
11.9 Summary 514
Problems 515
APPENDIX
A
Guide to Commands and Functions in
This Text 527
APPENDIX
B
Animation and Sound in MATLAB
APPENDIX
C
Formatted Output in MATLAB 549
APPENDIX
D
References
553
APPENDIX
E
Some Project Suggestions
www.mhhe.com/palm
Answers to Selected Problems 554
Index 557
538
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P R E FA C E
F
ormerly used mainly by specialists in signal processing and numerical
analysis, MATLAB® in recent years has achieved widespread and enthusiastic acceptance throughout the engineering community. Many engineering schools now require a course based entirely or in part on MATLAB early in
the curriculum. MATLAB is programmable and has the same logical, relational,
conditional, and loop structures as other programming languages, such as Fortran,
C, BASIC, and Pascal. Thus it can be used to teach programming principles. In
most schools a MATLAB course has replaced the traditional Fortran course, and
MATLAB is the principal computational tool used throughout the curriculum. In
some technical specialties, such as signal processing and control systems, it is
the standard software package for analysis and design.
The popularity of MATLAB is partly due to its long history, and thus it is
well developed and well tested. People trust its answers. Its popularity is also due
to its user interface, which provides an easy-to-use interactive environment that
includes extensive numerical computation and visualization capabilities. Its
compactness is a big advantage. For example, you can solve a set of many linear
algebraic equations with just three lines of code, a feat that is impossible with traditional programming languages. MATLAB is also extensible; currently more
than 20 “toolboxes” in various application areas can be used with MATLAB to
add new commands and capabilities.
MATLAB is available for MS Windows and Macintosh personal computers
and for other operating systems. It is compatible across all these platforms, which
enables users to share their programs, insights, and ideas. This text is based on
MATLAB version 7.9 (R2009b). Some of the material in Chapter 9 is based
on the control system toolbox, Version 8.4. Chapter 10 is based on Version 7.4 of
Simulink®. Chapter 11 is based on Version 5.3 of the Symbolic Math toolbox.
TEXT OBJECTIVES AND PREREQUISITES
This text is intended as a stand-alone introduction to MATLAB. It can be used in
an introductory course, as a self-study text, or as a supplementary text. The text’s
material is based on the author’s experience in teaching a required two-credit
semester course devoted to MATLAB for engineering freshmen. In addition,
the text can serve as a reference for later use. The text’s many tables and its
referencing system in an appendix have been designed with this purpose in mind.
A secondary objective is to introduce and reinforce the use of problemsolving methodology as practiced by the engineering profession in general and
®
MATLAB and Simulink are a registered trademarks of The MathWorks, Inc.
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Preface
as applied to the use of computers to solve problems in particular. This methodology is introduced in Chapter 1.
The reader is assumed to have some knowledge of algebra and trigonometry;
knowledge of calculus is not required for the rst seven chapters. Some knowledge of high school chemistry and physics, primarily simple electric circuits, and
basic statics and dynamics is required to understand some of the examples.
TEXT ORGANIZATION
This text is an update to the author’s previous text.* In addition to providing new
material based on MATLAB 7, especially the addition of the MuPAD program,
the text incorporates the many suggestions made by reviewers and other users.
The text consists of 11 chapters. The rst chapter gives an overview of
MATLAB features, including its windows and menu structures. It also introduces
the problem-solving methodology. Chapter 2 introduces the concept of an array,
which is the fundamental data element in MATLAB, and describes how to use numeric arrays, cell arrays, and structure arrays for basic mathematical operations.
Chapter 3 discusses the use of functions and les. MA TLAB has an extensive number of built-in math functions, and users can de ne their own functions
and save them as a le for reuse.
Chapter 4 treats programming with MATLAB and covers relational and logical operators, conditional statements, for and while loops, and the switch
structure. A major application of the chapter’s material is in simulation, to which
a section is devoted.
Chapter 5 treats two- and three-dimensional plotting. It rst establishes standards for professional-looking, useful plots. In the author’s experience, beginning
students are not aware of these standards, so they are emphasized. The chapter
then covers MATLAB commands for producing different types of plots and for
controlling their appearance.
Chapter 6 covers function discovery, which uses data plots to discover a
mathematical description of the data. It is a common application of plotting, and
a separate section is devoted to this topic. The chapter also treats polynomial and
multiple linear regression as part of its modeling coverage.
Chapter 7 reviews basic statistics and probability and shows how to use
MATLAB to generate histograms, perform calculations with the normal distribution, and create random number simulations. The chapter concludes with linear
and cubic spline interpolation. The following chapters are not dependent on the
material in this chapter.
Chapter 8 covers the solution of linear algebraic equations, which arise in applications in all elds of engineering. This coverage establishes the terminology
and some important concepts required to use the computer methods properly. The
chapter then shows how to use MATLAB to solve systems of linear equations
that have a unique solution. Underdetermined and overdetermined systems are
also covered. The remaining chapters are independent of this chapter.
*Introduction to MATLAB 7 for Engineers, McGraw-Hill, New York, 2005.
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Preface
Chapter 9 covers numerical methods for calculus and differential equations.
Numerical integration and differentiation methods are treated. Ordinary differential equation solvers in the core MATLAB program are covered, as well as the
linear system solvers in the Control System toolbox. This chapter provides some
background for Chapter 10.
Chapter 10 introduces Simulink, which is a graphical interface for building
simulations of dynamic systems. Simulink has increased in popularity and
has seen increased use in industry. This chapter need not be covered to read
Chapter 11.
Chapter 11 covers symbolic methods for manipulating algebraic expressions
and for solving algebraic and transcendental equations, calculus, differential
equations, and matrix algebra problems. The calculus applications include integration and differentiation, optimization, Taylor series, series evaluation, and
limits. Laplace transform methods for solving differential equations are also introduced. This chapter requires the use of the Symbolic Math toolbox, which includes MuPAD. MuPAD is a new feature in MATLAB. It provides a notebook
interface for entering commands and displaying results, including plots.
Appendix A contains a guide to the commands and functions introduced
in the text. Appendix B is an introduction to producing animation and sound
with MATLAB. While not essential to learning MATLAB, these features are
helpful for generating student interest. Appendix C summarizes functions for
creating formatted output. Appendix D is a list of references. Appendix E,
which is available on the text’s website, contains some suggestions for
course projects and is based on the author’s experience in teaching a freshman
MATLAB course. Answers to selected problems and an index appear at the
end of the text.
All gures, tables, equations, and exercises have been numbered according
to their chapter and section. For example, Figure 3.4–2 is the second gure in
Chapter 3, Section 4. This system is designed to help the reader locate these
items. The end-of-chapter problems are the exception to this numbering system.
They are numbered 1, 2, 3, and so on to avoid confusion with the in-chapter
exercises.
The rst four chapters constitute a course in the essentials of MA TLAB. The
remaining seven chapters are independent of one another, and may be covered in
any order or may be omitted if necessary. These chapters provide additional coverage and examples of plotting and model building, linear algebraic equations,
probability and statistics, calculus and differential equations, Simulink, and symbolic processing, respectively.
SPECIAL REFERENCE FEATURES
The text has the following special features, which have been designed to enhance
its usefulness as a reference.
■
Throughout each of the chapters, numerous tables summarize the commands and functions as they are introduced.
xi
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■
■
■
■
Appendix A is a complete summary of all the commands and functions
described in the text, grouped by category, along with the number of the
page on which they are introduced.
At the end of each chapter is a list of the key terms introduced in the
chapter, with the page number referenced.
Key terms have been placed in the margin or in section headings where
they are introduced.
The index has four sections: a listing of symbols, an alphabetical list of
MATLAB commands and functions, a list of Simulink blocks, and an
alphabetical list of topics.
PEDAGOGICAL AIDS
The following pedagogical aids have been included:
■
■
■
■
■
■
Each chapter begins with an overview.
Test Your Understanding exercises appear throughout the chapters near
the relevant text. These relatively straightforward exercises allow readers
to assess their grasp of the material as soon as it is covered. In most cases
the answer to the exercise is given with the exercise. Students should work
these exercises as they are encountered.
Each chapter ends with numerous problems, grouped according to the
relevant section.
Each chapter contains numerous practical examples. The major examples
are numbered.
Each chapter has a summary section that reviews the chapter’s objectives.
Answers to many end-of-chapter problems appear at the end of the text.
These problems are denoted by an asterisk next to their number (for
example, 15*).
Two features have been included to motivate the student toward MATLAB
and the engineering profession:
■
■
Most of the examples and the problems deal with engineering applications.
These are drawn from a variety of engineering elds and show realistic
applications of MATLAB. A guide to these examples appears on the inside
front cover.
The facing page of each chapter contains a photograph of a recent
engineering achievement that illustrates the challenging and interesting
opportunities that await engineers in the 21st century. A description of
the achievement and its related engineering disciplines and a discussion
of how MATLAB can be applied in those disciplines accompanies each
photo.
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Preface
ONLINE RESOURCES
An Instructor’s Manual is available online for instructors who have adopted this
text. This manual contains the complete solutions to all the Test Your Understanding exercises and to all the chapter problems. The text website (at
http://www.mhhe.com/palm) also has downloadable les containing PowerPoint
slides keyed to the text and suggestions for projects.
ELECTRONIC TEXTBOOK OPTIONS
Ebooks are an innovative way for students to save money and create a greener environment at the same time. An ebook can save students about one-half the cost of
a traditional textbook and offers unique features such as a powerful search engine,
highlighting, and the ability to share notes with classmates using ebooks.
McGraw-Hill offers this text as an ebook. To talk about the ebook options,
contact your McGraw-Hill sales rep or visit the site www.coursesmart.com to
learn more.
MATLAB INFORMATION
For MATLAB® and Simulink® product information, please contact:
The MathWorks, Inc.
3 Apple Hill Drive
Natick, MA, 01760-2098 USA
Tel: 508-647-7000
Fax: 508-647-7001
E-mail: info@mathworks.com
Web: www.mathworks.com
ACKNOWLEDGMENTS
Many individuals are due credit for this text. Working with faculty at the University of Rhode Island in developing and teaching a freshman course based on
MATLAB has greatly in uenced this text. Email from many users contained useful suggestions. The author greatly appreciates their contributions.
The MathWorks, Inc., has always been very supportive of educational publishing. I especially want to thank Naomi Fernandes of The MathWorks, Inc., for
her help. Bill Stenquist, Joyce Watters, and Lora Neyens of McGraw-Hill ef ciently handled the manuscript reviews and guided the text through production.
My sisters, Linda and Chris, and my mother, Lillian, have always been there,
cheering my efforts. My father was always there for support before he passed
away. Finally, I want to thank my wife, Mary Louise, and my children, Aileene,
Bill, and Andy, for their understanding and support of this project.
William J. Palm, III
Kingston, Rhode Island
September 2009
1
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Photo courtesy of NASA Jet Propulsion
Laboratory
Engineering in the
21st Century. . .
Remote Exploration
I
t will be many years before humans can travel to other planets. In the meantime, unmanned probes have been rapidly increasing our knowledge of the
universe. Their use will increase in the future as our technology develops to
make them more reliable and more versatile. Better sensors are expected for imaging and other data collection. Improved robotic devices will make these probes
more autonomous, and more capable of interacting with their environment, instead
of just observing it.
NASA’s planetary rover Sojourner landed on Mars on July 4, 1997, and excited people on Earth while they watched it successfully explore the Martian
surface to determine wheel-soil interactions, to analyze rocks and soil, and to
return images of the lander for damage assessment. Then in early 2004, two
improved rovers, Spirit and Opportunity, landed on opposite sides of the planet.
In one of the major discoveries of the 21st century, they obtained strong evidence
that water once existed on Mars in signi cant amounts.
About the size of a golf cart, the new rovers have six wheels, each with its
own motor. They have a top speed of 5 centimeters per second on at, hard
ground and can travel up to about 100 meters per day. Needing 100 watts to move,
they obtain power from solar arrays that generate 140 watts during a 4-hour
window each day. The sophisticated temperature control system must not only
protect against nighttime temperatures of 96C, but also prevent the rover from
overheating.
The robotic arm has three joints (shoulder, elbow, and wrist), driven by ve
motors, and it has a reach of 90 centimeters. The arm carries four tools and instruments for geological studies. Nine cameras provide hazard avoidance, navigation,
and panoramic views. The onboard computer has 128 MB of DRAM and coordinates all the subsystems including communications.
Although originally planned to last for three months, both rovers were still
exploring Mars at the end of 2009.
All engineering disciplines were involved with the rovers’ design and
launch. The MATLAB Neural Network, Signal Processing, Image Processing,
PDE, and various control system toolboxes are well suited to assist designers of
probes and autonomous vehicles like the Mars rovers. ■
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C H A P T E R
1
An Overview
of MATLAB®*
OUTLINE
1.1 MATLAB Interactive Sessions
1.2 Menus and the Toolbar
1.3 Arrays, Files, and Plots
1.4 Script Files and the Editor/Debugger
1.5 The MATLAB Help System
1.6 Problem-Solving Methodologies
1.7 Summary
Problems
This is the most important chapter in the book. By the time you have nished this
chapter, you will be able to use MATLAB to solve many kinds of problems.
Section 1.1 provides an introduction to MATLAB as an interactive calculator.
Section 1.2 covers the main menus and toolbar. Section 1.3 introduces arrays,
les, and plots. Section 1.4 discusses how to create, edit, and save MATLAB
programs. Section 1.5 introduces the extensive MATLAB Help System and
Section 1.6 introduces the methodology of engineering problem solving.
How to Use This Book
The book’s chapter organization is exible enough to accommodate a variety of
users. However, it is important to cover at least the rst four chapters, in that order.
Chapter 2 covers arrays, which are the basic building blocks in MATLAB. Chapter 3 covers le usage, functions built into MA TLAB, and user-de ned functions.
*MATLAB is a registered trademark of The MathWorks, Inc.
3
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CHAPTER 1
Page 4
An Overview of MATLAB®
Chapter 4 covers programming using relational and logical operators, conditional statements, and loops.
Chapters 5 through 11 are independent chapters that can be covered in any
order. They contain in-depth discussions of how to use MATLAB to solve several
common types of problems. Chapter 5 covers two- and three-dimensional plots in
greater detail. Chapter 6 shows how to use plots to build mathematical models
from data. Chapter 7 covers probability, statistics and interpolation applications.
Chapter 8 treats linear algebraic equations in more depth by developing methods
for the overdetermined and underdetermined cases. Chapter 9 introduces numerical methods for calculus and ordinary differential equations. Simulink®*, the topic
of Chapter 10, is a graphical user interface for solving differential equation
models. Chapter 11 covers symbolic processing with MuPAD®*, a new feature of
the MATLAB Symbolic Math toolbox, with applications to algebra, calculus,
differential equations, transforms, and special functions.
Reference and Learning Aids
The book has been designed as a reference as well as a learning tool. The special
features useful for these purposes are as follows.
■
■
■
■
■
■
■
Throughout each chapter margin notes identify where new terms are
introduced.
Throughout each chapter short Test Your Understanding exercises appear.
Where appropriate, answers immediately follow the exercise so you can
measure your mastery of the material.
Homework exercises conclude each chapter. These usually require greater
effort than the Test Your Understanding exercises.
Each chapter contains tables summarizing the MATLAB commands
introduced in that chapter.
At the end of each chapter is
■ A summary of what you should be able to do after completing that
chapter
■ A list of key terms you should know
Appendix A contains tables of MATLAB commands, grouped by category,
with the appropriate page references.
The index has four parts: MATLAB symbols, MATLAB commands,
Simulink blocks, and topics.
1.1 MATLAB Interactive Sessions
We now show how to start MATLAB, how to make some basic calculations, and
how to exit MATLAB.
*Simulink and MuPAD are registered trademarks of The MathWorks, Inc.
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1.1
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Figure 1.1–1 The default MATLAB Desktop.
Conventions
In this text we use typewriter font to represent MATLAB commands, any
text that you type in the computer, and any MATLAB responses that appear on
the screen, for example, y = 6*x. Variables in normal mathematics text appear
in italics, for example, y 6x. We use boldface type for three purposes: to represent vectors and matrices in normal mathematics text (for example, Ax ⴝ b), to
represent a key on the keyboard (for example, Enter), and to represent the name
of a screen menu or an item that appears in such a menu (for example, File). It is
assumed that you press the Enter key after you type a command. We do not show
this action with a separate symbol.
Starting MATLAB
To start MATLAB on a MS Windows system, double-click on the MATLAB icon.
You will then see the MATLAB Desktop. The Desktop manages the Command
window and a Help Browser as well as other tools. The default appearance of the
Desktop is shown in Figure 1.1–1. Five windows appear. These are the Command
window in the center, the Command History window in the lower right, the
Workspace window in the upper right, the Details window in the lower left, and the
DESKTOP
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Current Directory window in the upper left. Across the top of the Desktop are a row
of menu names and a row of icons called the toolbar. To the right of the toolbar is
a box showing the directory where MATLAB looks for and saves les. We will
describe the menus, toolbar, and directories later in this chapter.
You use the Command window to communicate with the MATLAB program, by typing instructions of various types called commands, functions, and
statements. Later we will discuss the differences between these types, but for
now, to simplify the discussion, we will call the instructions by the generic name
commands. MATLAB displays the prompt (>>) to indicate that it is ready to
receive instructions. Before you give MATLAB instructions, make sure the cursor is located just after the prompt. If it is not, use the mouse to move the cursor.
The prompt in the Student Edition looks like EDU >>. We will use the normal
prompt symbol >> to illustrate commands in this text. The Command window in
Figure 1.1–1 shows some commands and the results of the calculations. We will
cover these commands later in this chapter.
Four other windows appear in the default Desktop. The Current Directory
window is much like a le manager window; you can use it to access les.
Double-clicking on a le name with the extension .m will open that le in the
MATLAB Editor. The Editor is discussed in Section 1.4. Figure 1.1–1 shows
the les in the author ’s directory C:\MyMATLABFiles.
Underneath the Current Directory window is the . . . window. It displays any
comments in the le. Note that two le types are shown in the Current Directory .
These have the extensions .m and .mdl. We will cover M les in this chapter .
Chapter 10 covers Simulink, which uses MDL les. You can have other le types
in the directory.
The Workspace window appears in the upper right. The Workspace window
displays the variables created in the Command window. Double-click on a variable name to open the Array Editor, which is discussed in Chapter 2.
The fth window in the default Desktop is the Command History window .
This window shows all the previous keystrokes you entered in the Command
window. It is useful for keeping track of what you typed. You can click on a
keystroke and drag it to the Command window or the Editor to avoid retyping it.
Double-clicking on a keystroke executes it in the Command window.
You can alter the appearance of the Desktop if you wish. For example, to
eliminate a window, just click on its Close-window button () in its upper righthand corner. To undock, or separate the window from the Desktop, click on the
button containing a curved arrow. An undocked window can be moved around on
the screen. You can manipulate other windows in the same way. To restore the
default con guration, click on the Desktop menu, then click on Desktop Layout,
and select Default.
Entering Commands and Expressions
To see how simple it is to use MATLAB, try entering a few commands on your
computer. If you make a typing mistake, just press the Enter key until you get
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the prompt, and then retype the line. Or, because MATLAB retains your previous
keystrokes in a command le, you can use the up-arrow key ( ) to scroll back
through the commands. Press the key once to see the previous entry, twice to
see the entry before that, and so on. Use the down-arrow key (↓) to scroll forward
through the commands. When you nd the line you want, you can edit it using
the left- and right-arrow keys (← and →), and the Backspace key, and the Delete
key. Press the Enter key to execute the command. This technique enables you to
correct typing mistakes quickly.
Note that you can see your previous keystrokes displayed in the Command
History window. You can copy a line from this window to the Command window
by highlighting the line with the mouse, holding down the left mouse button, and
dragging the line to the Command window.
Make sure the cursor is at the prompt in the Command window. To divide
8 by 10, type 8/10 and press Enter (the symbol / is the MATLAB symbol for
division). Your entry and the MATLAB response look like the following on
the screen (we call this interaction between you and MATLAB an interactive
session, or simply a session). Remember, the symbol >> automatically appears
on the screen; you do not type it.
7
↓
SESSION
>> 8/10
ans =
0.8000
MATLAB indents the numerical result. MATLAB uses high precision for its
computations, but by default it usually displays its results using four decimal
places except when the result is an integer.
MATLAB assigns the most recent answer to a variable called ans, which is
an abbreviation for answer. A variable in MATLAB is a symbol used to contain
a value. You can use the variable ans for further calculations; for example, using
the MATLAB symbol for multiplication (*), we obtain
>> 5*ans
ans =
4
Note that the variable ans now has the value 4.
You can use variables to write mathematical expressions. Instead of using
the default variable ans, you can assign the result to a variable of your own
choosing, say, r, as follows:
>> r=8/10
r =
0.8000
Spaces in the line improve its readability; for example, you can put a space
before and after the = sign if you want. MATLAB ignores these spaces when
making its calculations. It also ignores spaces surrounding and signs.
VARIABLE
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If you now type r at the prompt and press Enter, you will see
>> r
r =
0.8000
thus verifying that the variable r has the value 0.8. You can use this variable in
further calculations. For example,
>> s=20*r
s =
16
ARGUMENT
A common mistake is to forget the multiplication symbol * and type the expression as you would in algebra, as s 20r. If you do this in MATLAB, you
will get an error message.
MATLAB has hundreds of functions available. One of these is the square
root function, sqrt. A pair of parentheses is used after the function’s name to
enclose the value—called the function’s argument—that is operated on by the
function. For example, to compute the square root of 9 and assign its value to
the variable r, you type r = sqrt(9). Note that the previous value of r has
been replaced by 3.
Order of Precedence
SCALAR
A scalar is a single number. A scalar variable is a variable that contains a single
number. MATLAB uses the symbols * / ^ for addition, subtraction,
multiplication, division, and exponentiation (power) of scalars. These are listed
in Table 1.1–1. For example, typing x = 8 + 3*5 returns the answer x = 23.
Typing 2^3-10 returns the answer ans = -2. The forward slash (/ ) represents right division, which is the normal division operator familiar to you.
Typing 15/3 returns the result ans = 5.
MATLAB has another division operator, called left division, which is denoted by the backslash (\). The left division operator is useful for solving sets of
linear algebraic equations, as we will see. A good way to remember the difference between the right and left division operators is to note that the slash slants
toward the denominator. For example, 7/2 2\7 3.5.
Table 1.1–1 Scalar arithmetic operations
Symbol
Operation
MATLAB form
b
^
*
/
exponentiation: a
multiplication: ab
right division: a/b ba
a^b
a*b
a/b
\
left division: a\b ba
a\b
addition: a b
subtraction: a b
ab
ab
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Table 1.1–2 Order of precedence
Precedence
Operation
First
Second
Third
Parentheses, evaluated starting with the innermost pair.
Exponentiation, evaluated from left to right.
Multiplication and division with equal precedence, evaluated from
left to right.
Addition and subtraction with equal precedence, evaluated from
left to right.
Fourth
The mathematical operations represented by the symbols * / \ and
^ follow a set of rules called precedence. Mathematical expressions are evaluated
starting from the left, with the exponentiation operation having the highest order of
precedence, followed by multiplication and division with equal precedence, followed by addition and subtraction with equal precedence. Parentheses can be used
to alter this order. Evaluation begins with the innermost pair of parentheses and
proceeds outward. Table 1.1–2 summarizes these rules. For example, note the
effect of precedence on the following session.
>>8 + 3*5
ans =
23
>>(8 + 3)*5
ans =
55
>>4^2 - 12 - 8/4*2
ans =
0
>>4^2 - 12 - 8/(4*2)
ans =
3
>>3*4^2 + 5
ans =
53
>>(3*4)^2 + 5
ans =
149
>>27^(1/3) + 32^(0.2)
ans =
5
>>27^(1/3) + 32^0.2
ans =
5
>>27^1/3 + 32^0.2
ans =
11
PRECEDENCE
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To avoid mistakes, feel free to insert parentheses wherever you are unsure of the
effect precedence will have on the calculation. Use of parentheses also improves
the readability of your MATLAB expressions. For example, parentheses are not
needed in the expression 8+(3*5), but they make clear our intention to multiply 3 by 5 before adding 8 to the result.
Test Your Understanding
T1.1–1 Use MATLAB to compute the following expressions.
a. 6 a
10
18
b +
+ 5(92)
13
5(7)
b. 6(351/4) + 140.35
(Answers: a. 410.1297 b. 17.1123.)
The Assignment Operator
The = sign in MATLAB is called the assignment or replacement operator. It works
differently than the equals sign you know from mathematics. When you type
x = 3, you tell MATLAB to assign the value 3 to the variable x. This usage is no
different than in mathematics. However, in MATLAB we can also type something
like this: x = x + 2. This tells MATLAB to add 2 to the current value of x, and
to replace the current value of x with this new value. If x originally had the value 3,
its new value would be 5. This use of the operator is different from its use in
mathematics. For example, the mathematics equation x x 2 is invalid because
it implies that 0 2.
In MATLAB the variable on the left-hand side of the = operator is replaced
by the value generated by the right-hand side. Therefore, one variable, and only
one variable, must be on the left-hand side of the = operator. Thus in MATLAB
you cannot type 6 = x. Another consequence of this restriction is that you
cannot write in MATLAB expressions like the following:
>>x+2=20
The corresponding equation x 2 20 is acceptable in algebra and has the solution x 18, but MATLAB cannot solve such an equation without additional
commands (these commands are available in the Symbolic Math toolbox, which
is described in Chapter 11).
Another restriction is that the right-hand side of the = operator must have a
computable value. For example, if the variable y has not been assigned a value,
then the following will generate an error message in MATLAB.
>>x = 5 + y
In addition to assigning known values to variables, the assignment operator
is very useful for assigning values that are not known ahead of time, or for
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changing the value of a variable by using a prescribed procedure. The following
example shows how this is done.
Volume of a Circular Cylinder
EXAMPLE 1.1–1
The volume of a circular cylinder of height h and radius r is given by V ⫽ ␲r2h. A particular cylindrical tank is 15 m tall and has a radius of 8 m. We want to construct another
cylindrical tank with a volume 20 percent greater but having the same height. How large
must its radius be?
■ Solution
First solve the cylinder equation for the radius r. This gives
r =
V
␲h
B
The session is shown below. First we assign values to the variables r and h representing the
radius and height. Then we compute the volume of the original cylinder and increase
the volume by 20 percent. Finally we solve for the required radius. For this problem we can
use the MATLAB built-in constant pi.
>>r = 8;
>>h = 15;
>>V = pi*r^2*h;
>>V = V + 0.2*V;
>>r = sqrt(V/(pi*h))
r =
8.7636
Thus the new cylinder must have a radius of 8.7636 m. Note that the original values of
the variables r and V are replaced with the new values. This is acceptable as long as we
do not wish to use the original values again. Note how precedence applies to the line V =
pi*r^2*h;. It is equivalent to V = pi*(r^2)*h;.
Variable Names
The term workspace refers to the names and values of any variables in use in the
current work session. Variable names must begin with a letter; the rest of the
name can contain letters, digits, and underscore characters. MATLAB is casesensitive. Thus the following names represent ve dif ferent variables: speed,
Speed, SPEED, Speed_1, and Speed_2. In MATLAB 7, variable names
can be no longer than 63 characters.
Managing the Work Session
Table 1.1–3 summarizes some commands and special symbols for managing the
work session. A semicolon at the end of a line suppresses printing the results to
the screen. If a semicolon is not put at the end of a line, MATLAB displays the
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Table 1.1–3 Commands for managing the work session
Command
Description
clc
clear
clear var1 var2
exist(‘name’)
quit
who
whos
Clears the Command window.
Removes all variables from memory.
Removes the variables var1 and var2 from memory.
Determines if a le or variable exists having the name ‘name’.
Stops MATLAB.
Lists the variables currently in memory.
Lists the current variables and sizes and indicate if they have
imaginary parts.
Colon; generates an array having regularly spaced elements.
Comma; separates elements of an array.
Semicolon; suppresses screen printing; also denotes a new row
in an array.
Ellipsis; continues a line.
:
,
;
...
results of the line on the screen. Even if you suppress the display with the semicolon, MATLAB still retains the variable’s value.
You can put several commands on the same line if you separate them with a
comma if you want to see the results of the previous command or semicolon if
you want to suppress the display. For example,
>>x=2;y=6+x,x=y+7
y =
8
x =
15
Note that the rst value of x was not displayed. Note also that the value of x
changed from 2 to 15.
If you need to type a long line, you can use an ellipsis, by typing three
periods, to delay execution. For example,
>>NumberOfApples = 10; NumberOfOranges = 25;
>>NumberOfPears = 12;
>>FruitPurchased = NumberOfApples + NumberOfOranges ...
+NumberOfPears
FruitPurchased =
47
Use the arrow, Tab, and Ctrl keys to recall, edit, and reuse functions and
variables you typed earlier. For example, suppose you mistakenly enter the line
>>volume = 1 + sqr(5)
MATLAB responds with an error message because you misspelled sqrt.
Instead of retyping the entire line, press the up-arrow key ( ) once to display
the previously typed line. Press the left-arrow key (←) several times to move
the cursor and add the missing t, then press Enter. Repeated use of the up-arrow
key recalls lines typed earlier.
↓
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Tab and Arrow Keys
You can use the smart recall feature to recall a previously typed function or variable whose rst few characters you specify . For example, after you have entered
the line starting with volume, typing vol and pressing the up-arrow key ( )
once recalls the last-typed line that starts with the function or variable whose
name begins with vol. This feature is case-sensitive.
You can use the tab completion feature to reduce the amount of typing.
MATLAB automatically completes the name of a function, variable, or le if
you type the rst few letters of the name and press the Tab key. If the name is
unique, it is automatically completed. For example, in the session listed earlier, if
you type Fruit and press Tab, MATLAB completes the name and displays
FruitPurchased. Press Enter to display the value of the variable, or continue
editing to create a new executable line that uses the variable FruitPurchased.
If there is more than one name that starts with the letters you typed, MATLAB
displays these names when you press the Tab key. Use the mouse to select the
desired name from the pop-up list by double-clicking on its name.
The left-arrow (←) and right-arrow (→) keys move left and right through
a line one character at a time. To move through one word at a time, press Ctrl
and → simultaneously to move to the right; press Ctrl and ← simultaneously
to move to the left. Press Home to move to the beginning of a line; press End
to move to the end of a line.
↓
Deleting and Clearing
Press Del to delete the character at the cursor; press Backspace to delete the character before the cursor. Press Esc to clear the entire line; press Ctrl and k simultaneously to delete (kill) to the end of the line.
MATLAB retains the last value of a variable until you quit MATLAB or clear
its value. Overlooking this fact commonly causes errors in MATLAB. For example, you might prefer to use the variable x in a number of different calculations. If
you forget to enter the correct value for x, MATLAB uses the last value, and you
get an incorrect result. You can use the clear function to remove the values of
all variables from memory, or you can use the form clear var1 var2 to clear
the variables named var1 and var2. The effect of the clc command is different; it clears the Command window of everything in the window display, but the
values of the variables remain.
You can type the name of a variable and press Enter to see its current value.
If the variable does not have a value (i.e., if it does not exist), you see an error
message. You can also use the exist function. Type exist(‘x’) to see if the
variable x is in use. If a 1 is returned, the variable exists; a 0 indicates that it does
not exist. The who function lists the names of all the variables in memory, but
does not give their values. The form who var1 var2 restricts the display to the
variables speci ed. The wildcard character * can be used to display variables that
match a pattern. For instance, who A* nds all variables in the current
workspace that start with A. The whos function lists the variable names and their
sizes and indicates whether they have nonzero imaginary parts.
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The difference between a function and a command or a statement is that functions have their arguments enclosed in parentheses. Commands, such as clear,
need not have arguments; but if they do, they are not enclosed in parentheses, for
example, clear x. Statements cannot have arguments; for example, clc and
quit are statements.
Press Ctrl-C to cancel a long computation without terminating the session.
You can quit MATLAB by typing quit. You can also click on the File menu,
and then click on Exit MATLAB.
Prede ned Constants
MATLAB has several prede ned special constants, such as the built-in constant
pi we used in Example 1.1–1. Table 1.1–4 lists them. The symbol Inf stands
for ⬁, which in practice means a number so large that MATLAB cannot represent it. For example, typing 5/0 generates the answer Inf. The symbol NaN
stands for “not a number.” It indicates an unde ned numerical result such as that
obtained by typing 0/0. The symbol eps is the smallest number which, when
added to 1 by the computer, creates a number greater than 1.We use it as an indicator of the accuracy of computations.
The symbols i and j denote the imaginary unit, where i = j = 1 - 1. We use
them to create and represent complex numbers, such as x = 5 + 8i.
Try not to use the names of special constants as variable names. Although
MATLAB allows you to assign a different value to these constants, it is not good
practice to do so.
Complex Number Operations
MATLAB handles complex number algebra automatically. For example, the
number c1 ⫽ 1 ⫺ 2i is entered as follows: c1 = 1-2i. You can also type c1 =
Complex(1, -2).
Caution: Note that an asterisk is not needed between i or j and a number, although
it is required with a variable, such as c2 = 5 - i*c1. This convention can cause
errors if you are not careful. For example, the expressions y = 7/2*i and x =
7/2i give two different results: y ⫽ (7/2)i ⫽ 3.5i and x ⫽ 7/(2i) ⫽ ⫺3.5i.
Table 1.1–4 Special variables and constants
Command
Description
ans
eps
i,j
Inf
NaN
pi
Temporary variable containing the most recent answer.
Speci es the accuracy of oating point precision.
The imaginary unit 1-1.
In nity .
Indicates an unde ned numerical result.
The number ␲.
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Addition, subtraction, multiplication, and division of complex numbers are
easily done. For example,
>>s = 3+7i;w = 5-9i;
>>w+s
ans =
8.0000 - 2.0000i
>>w*s
ans =
78.0000 + 8.0000i
>>w/s
ans =
-0.8276 - 1.0690i
Test Your Understanding
T1.1–2 Given x 5 9i and y 6 2i, use MATLAB to show that x y 1 7i, xy 12 64i, and x/y 1.2 1.1i.
Formatting Commands
The format command controls how numbers appear on the screen. Table 1.1–5
gives the variants of this command. MATLAB uses many signi cant gures in its
calculations, but we rarely need to see all of them. The default MATLAB display
format is the short format, which uses four decimal digits. You can display more
by typing format long, which gives 16 digits. To return to the default mode,
type format short.
You can force the output to be in scienti c notation by typing format
short e, or format long e, where e stands for the number 10. Thus the output 6.3792e+03 stands for the number 6.3792 103. The output 6.3792e-03
Table 1.1–5 Numeric display formats
Command
Description and example
format short
format long
format short e
Four decimal digits (the default); 13.6745.
16 digits; 17.27484029463547.
Five digits (four decimals) plus exponent;
6.3792e03.
16 digits (15 decimals) plus exponent;
6.379243784781294e04.
Two decimal digits; 126.73.
Positive, negative, or zero; .
Rational approximation; 43/7.
Suppresses some blank lines.
Resets to less compact display mode.
format long e
format bank
format format rat
format compact
format loose
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Figure 1.2–1 The top of the MATLAB Desktop.
stands for the number 6.3792 103. Note that in this context e does not
represent the number e, which is the base of the natural logarithm. Here e stands
for “exponent.” It is a poor choice of notation, but MATLAB follows conventional
computer programming standards that were established many years ago.
Use format bank only for monetary calculations; it does not recognize
imaginary parts.
1.2 Menus and the Toolbar
CURRENT
DIRECTORY
The Desktop manages the Command window and other MATLAB tools. The
default appearance of the Desktop is shown in Figure 1.1–1. Across the top of
the Desktop are a row of menu names and a row of icons called the toolbar. To
the right of the toolbar is a box showing the current directory, where MATLAB
looks for les. See Figure 1.2–1.
Other windows appear in a MATLAB session, depending on what you do.
For example, a graphics window containing a plot appears when you use the
plotting functions; an editor window, called the Editor/Debugger, appears for use
in creating program les. Each window type has its own menu bar , with one or
more menus, at the top. Thus the menu bar will change as you change windows.
To activate or select a menu, click on it. Each menu has several items. Click on
an item to select it. Keep in mind that menus are context-sensitive. Thus their
contents change, depending on which features you are currently using.
The Desktop Menus
Most of your interaction will be in the Command window. When the Command
window is active, the default MATLAB 7 Desktop (shown in Figure 1.1–1) has
six menus: File, Edit, Debug, Desktop, Window, and Help. Note that these
menus change depending on what window is active. Every item on a menu can
be selected with the menu open either by clicking on the item or by typing its
underlined letter. Some items can be selected without the menu being open by
using the shortcut key listed to the right of the item. Those items followed by
three dots (. . .) open a submenu or another window containing a dialog box.
The three most useful menus are the File, Edit, and Help menus. The Help
menu is described in Section 1.5. The File menu in MATLAB 7 contains the following items, which perform the indicated actions when you select them.
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The File Menu in MATLAB 7
New Opens a dialog box that allows you to create a new program le, called
an M- le, using a text editor called the Editor/Debugger , a new Figure,
a variable in the Workspace window, Model le (a le type used by
Simulink), or a new GUI (which stands for Graphical User Interface).
Open. . . Opens a dialog box that allows you to select a le for editing.
Close Command Window (or Current Folder) Closes the Command
window or current le if one is open.
Import Data. . . Starts the Import Wizard which enables you to import data
easily.
Save Workspace As. . . Opens a dialog box that enables you to save a le.
Set Path. . . Opens a dialog box that enables you to set the MATLAB search
path.
Preferences. . . Opens a dialog box that enables you to set preferences for
such items as fonts, colors, tab spacing, and so forth.
Page Setup Opens a dialog box that enables you to format printed output.
Print. . . Opens a dialog box that enables you to print all the Command
window.
Print Selection. . . Opens a dialog box that enables you to print selected
portions of the Command window.
File List Contains a list of previously used les, in order of most recently
used.
Exit MATLAB Closes MATLAB.
The New option in the File menu lets you select which type of M- le to
create: a blank M- le, a function M- le, or a class M- le. Select blank M- le
to create an M- le of the type discussed in Section 1.4. Function M- les are discussed in Chapter 3, but class M- les are beyond the scope of this text.
The Edit menu contains the following items.
The Edit Menu in MATLAB 7
Undo Reverses the previous editing operation.
Redo Reverses the previous Undo operation.
Cut Removes the selected text and stores it for pasting later.
Copy Copies the selected text for pasting later, without removing it.
Paste Inserts any text on the clipboard at the current location of the cursor.
Paste to Workspace. . . Inserts the contents of the clipboard into the
workspace as one or more variables.
Select All Highlights all text in the Command window.
Delete Clears the variable highlighted in the Workspace Browser.
Find. . . Finds and replaces phrases.
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Find Files. . . Finds les.
Clear Command Window Removes all text from the Command window.
Clear Command History Removes all text from the Command History
window.
Clear Workspace Removes the values of all variables from the workspace.
You can use the Copy and Paste selections to copy and paste commands appearing
on the Command window. However, an easier way is to use the up-arrow key to
scroll through the previous commands, and press Enter when you see the command
you want to retrieve.
Use the Debug menu to access the Debugger, which is discussed in Chapter 4.
Use the Desktop menu to control the con guration of the Desktop and to display
toolbars. The Window menu has one or more items, depending on what you
have done thus far in your session. Click on the name of a window that appears
on the menu to open it. For example, if you have created a plot and not closed its
window, the plot window will appear on this menu as Figure 1. However, there
are other ways to move between windows (such as pressing the Alt and Tab keys
simultaneously if the windows are not docked).
The View menu will appear to the right of the Edit menu if you have selected a le in the folder in the Current Folder window . This menu gives information about the selected le.
The toolbar, which is below the menu bar, provides buttons as shortcuts to
some of the features on the menus. Clicking on the button is equivalent to clicking on the menu, then clicking on the menu item; thus the button eliminates one
click of the mouse. The rst seven buttons from the left correspond to the New
M-File, Open File, Cut, Copy, Paste, Undo, and Redo. The eighth button activates Simulink, which is a program built on top of MATLAB. The ninth button
activates the GUIDE Quick Start, which is used to create and edit graphical user
interfaces (GUIs). The tenth button activates the Pro ler , which can be used to
optimize program performance. The eleventh button (the one with the question
mark) accesses the Help System.
Below the toolbar is a button that accesses help for adding shortcuts to the toolbar and a button that accesses a list of the features added since the previous release.
1.3 Arrays, Files, and Plots
This section introduces arrays, which are the basic building blocks in MATLAB,
and shows how to handle les and generate plots.
Arrays
MATLAB has hundreds of functions, which we will discuss throughout the text.
For example, to compute sin x, where x has a value in radians, you type sin(x).
To compute cos x, type cos(x). The exponential function ex is computed from
exp(x). The natural logarithm, ln x, is computed by typing log(x). (Note the
spelling difference between mathematics text, ln, and MATLAB syntax, log.)
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You compute the base-10 logarithm by typing log10(x). The inverse sine, or
arcsine, is obtained by typing asin(x). It returns an answer in radians, not
degrees. The function asind(x) returns degrees.
One of the strengths of MATLAB is its ability to handle collections of numbers, called arrays, as if they were a single variable. A numerical array is an ordered collection of numbers (a set of numbers arranged in a speci c order). An
example of an array variable is one that contains the numbers 0, 4, 3, and 6, in
that order. We use square brackets to de ne the variable x to contain this collection by typing x = [0, 4, 3, 6]. The elements of the array may also be
separated by spaces, but commas are preferred to improve readability and avoid
mistakes. Note that the variable y de ned as y = [6, 3, 4, 0] is not the
same as x because the order is different. The reason for using the brackets is as
follows. If you were to type x = 0, 4, 3, 6, MATLAB would treat this as
four separate inputs and would assign the value 0 to x. The array [0, 4, 3, 6]
can be considered to have one row and four columns, and it is a subcase of a
matrix, which has multiple rows and columns. As we will see, matrices are also
denoted by square brackets.
We can add the two arrays x and y to produce another array z by typing the
single line z = x + y. To compute z, MATLAB adds all the corresponding numbers in x and y to produce z. The resulting array z contains the numbers 6, 7, 7, 6.
You need not type all the numbers in the array if they are regularly spaced.
Instead, you type the rst number and the last number , with the spacing in the
middle, separated by colons. For example, the numbers 0, 0.1, 0.2, . . . , 10 can
be assigned to the variable u by typing u = 0:0.1:10. In this application of
the colon operator, the brackets should not be used.
To compute w 5 sin u for u 0, 0.1, 0.2 , . . . , 10, the session is
19
ARRAY
>>u = 0:0.1:10;
>>w = 5*sin(u);
The single line w = 5*sin(u) computed the formula w 5 sin u 101 times,
once for each value in the array u, to produce an array z that has 101 values.
You can see all the u values by typing u after the prompt; or, for example,
you can see the seventh value by typing u(7). The number 7 is called an array
index, because it points to a particular element in the array.
>>u(7)
ans =
0.6000
>>w(7)
ans =
2.8232
You can use the length function to determine how many values are in an
array. For example, continue the previous session as follows:
>>m = length(w)
m =
101
ARRAY INDEX
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Arrays that display on the screen as a single row of numbers with more than
one column are called row arrays. You can create column arrays, which have
more than one row, by using a semicolon to separate the rows.
Polynomial Roots
We can describe a polynomial in MATLAB with an array whose elements are the
polynomial’s coef cients, starting with the coef cient of the highest power of x.
For example, the polynomial 4x3 8x2 7x 5 would be represented by the
array[4,-8,7,-5]. The roots of the polynomial f (x) are the values of x such that
f (x) 0. Polynomial roots can be found with the roots(a) function, where a is
the polynomial’s coef cient array. The result is a column array that contains the
polynomial’s roots. For example, to nd the roots of x3 7x2 40x 34 0,
the session is
>>a = [1,-7,40,-34];
>>roots(a)
ans =
3.0000 + 5.000i
3.0000 - 5.000i
1.0000
The roots are x 1 and x 3 5i. The two commands could have been combined into the single command roots([1,-7,40,-34]).
Test Your Understanding
T1.3–1 Use MATLAB to determine how many elements are in the array
cos(0):0.02:log10(100). Use MATLAB to determine the
25th element. (Answer: 51 elements and 1.48.)
T1.3–2 Use MATLAB to nd the roots of the polynomial 290 11x 6x2 x3.
(Answer: x 10, 2 5i.)
Built-in Functions
We have seen several of the functions built into MATLAB, such as the sqrt and
sin functions. Table 1.3–1 lists some of the commonly used built-in functions.
Chapter 3 gives extensive coverage of the built-in functions. MATLAB users can
create their own functions for their special needs. Creation of user-de ned functions
is covered in Chapter 3.
Working with Files
MAT-FILES
MATLAB uses several types of les that enable you to save programs, data, and
session results. As we will see in Section 1.4, MATLAB function les and program les are saved with the extension . m, and thus are called M- les. MAT- les
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Table 1.3–1 Some commonly used mathematical functions
Function
MATLAB syntax*
x
exp (x)
sqrt (x)
log (x)
log 10(x)
cos (x)
sin (x)
tan (x)
acos (x)
asin (x)
atan (x)
e
1x
ln x
log10 x
cos x
sin x
tan x
cos1 x
sin1 x
tan1 x
*The MATLAB trigonometric functions listed here use radian measure. Trigonometric functions ending
in d, such as sind(x) and cosd(x), take the argument x in degrees. Inverse functions such as
atand(x) return values in degrees.
have the extension .mat and are used to save the names and values of variables
created during a MATLAB session.
Because they are ASCII les, M- les can be created using just about any
word processor. MAT- les are binary les that are generally readable only by
the software that created them. MAT- les contain a machine signature that
allows them to be transferred between machine types such as MS Windows and
Macintosh machines.
The third type of file we will be using is a data file, specifically an ASCII
data file, that is, one created according to the ASCII format. You may need to
use MATLAB to analyze data stored in such a file created by a spreadsheet
program, a word processor, or a laboratory data acquisition system or in a file
you share with someone else.
Saving and Retrieving Your Workspace Variables
If you want to continue a MATLAB session at a later time, you must use the save
and load commands. Typing save causes MATLAB to save the workspace
variables, that is, the variable names, their sizes, and their values, in a binary
le called matlab.mat, which MATLAB can read. To retrieve your
workspace variables, type load. You can then continue your session as before.
To save the workspace variables in another le named lename.mat, type
save lename. To load the workspace variables, type load lename. If
the saved MAT- le lename contains the variables A, B, and C, then loading the le lename places these variables back into the workspace and overwrites any existing variables having the same name.
To save just some of your variables, say, var1 and var2, in the le
lename.mat, type save lename var1 var2. You need not type the
variable names to retrieve them; just type load lename.
Directories and Path It is important to know the location of the les you use
with MATLAB. File location frequently causes problems for beginners. Suppose
ASCII FILES
DATA FILE
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you use MATLAB on your home computer and save a le to a removable disk, as
discussed later in this section. If you bring that disk to use with MATLAB on another computer, say, in a school’s computer lab, you must make sure that MATLAB
knows how to nd your les. Files are stored in directories, called folders on some
computer systems. Directories can have subdirectories below them. For example,
suppose MATLAB was installed on drive c: in the directory c:\matlab. Then
the toolbox directory is a subdirectory under the directory c:\matlab, and
symbolic is a subdirectory under the toolbox directory. The path tells us and
MATLAB how to nd a particular le.
Working with Removable Disks In Section 1.4 you will learn how to create
and save M- les. Suppose you have saved the le problem1.m in the directory
\homework on a disk, which you insert in drive f:. The path for this
le is f:\homework. As MATLAB is normally installed, when you type
problem1,
SEARCH PATH
1. MATLAB rst checks to see if problem1 is a variable and if so, displays
its value.
2. If not, MATLAB then checks to see if problem1 is one of its own
commands, and executes it if it is.
3. If not, MATLAB then looks in the current directory for a le named
problem1.m and executes problem1 if it nds it.
4. If not, MATLAB then searches the directories in its search path, in order,
for problem1.m and then executes it if found.
You can display the MATLAB search path by typing path. If problem1 is on
the disk only and if directory f: is not in the search path, MATLAB will not nd
the le and will generate an error message, unless you tell it where to look. You
can do this by typing cd f:\homework, which stands for “change directory
to f:\homework.” This will change the current directory to f:\homework and
force MATLAB to look in that directory to nd your le. The general syntax
of this command is cd dirname, where dirname is the full path to the
directory.
An alternative to this procedure is to copy your le to a directory on the hard
drive that is in the search path. However, there are several pitfalls with this approach:
(1) if you change the le during your session, you might forget to copy the revised le
back to your disk; (2) the hard drive becomes cluttered (this is a problem in public
computer labs, and you might not be permitted to save your le on the hard drive);
(3) the le might be deleted or overwritten if MATLAB is reinstalled; and (4) someone else can access your work!
You can determine the current directory (the one where MATLAB looks for
your le) by typing pwd. To see a list of all the les in the current directory , type
dir. To see the les in the directory dirname, type dir dirname.
The what command displays a list of the MATLAB-speci c les in the current directory. The what dirname command does the same for the directory
dirname. Type which item to display the full path name of the function
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Table 1.3–2 System, directory, and le commands
Command
Description
addpath dirname
cd dirname
dir
dir dirname
path
pathtool
pwd
rmpath dirname
what
Adds the directory dirname to the search path.
Changes the current directory to dirname.
Lists all les in the current directory .
Lists all the les in the directory dirname.
Displays the MATLAB search path.
Starts the Set Path tool.
Displays the current directory.
Removes the directory dirname from the search path.
Lists the MATLAB-speci c les found in the current
working directory. Most data les and other non-MA TLAB
les are not listed. Use dir to get a list of all les.
Lists the MATLAB-speci c les in directory dirname.
Displays the path name of item if item is a function or
le. Identi es item as a variable if so.
what dirname
which item
item or the le item (include the le extension). If item is a variable, then
MATLAB identi es it as such.
You can add a directory to the search path by using the addpath command.
To remove a directory from the search path, use the rmpath command. The Set
Path tool is a graphical interface for working with les and directories. Type
pathtool to start the browser. To save the path settings, click on Save in the
tool. To restore the default search path, click on Default in the browser.
These commands are summarized in Table 1.3–2.
Plotting with MATLAB
MATLAB contains many powerful functions for easily creating plots of several
different types, such as rectilinear, logarithmic, surface, and contour plots. As a
simple example, let us plot the function y 5 sin x for 0 x 7. We choose to
use an increment of 0.01 to generate a large number of x values in order to
produce a smooth curve. The function plot(x,y) generates a plot with the
x values on the horizontal axis (the abscissa) and the y values on the vertical axis
(the ordinate). The session is
>>x = 0:0.01:7;
>>y = 3*cos(2*x);
>>plot(x,y),xlabel(‘x’),ylabel(‘y’)
The plot appears on the screen in a graphics window, named Figure 1, as
shown in Figure 1.3–1. The xlabel function places the text in single quotes
as a label on the horizontal axis. The ylabel function performs a similar
function for the vertical axis. When the plot command is successfully executed,
a graphics window automatically appears. If a hard copy of the plot is desired,
GRAPHICS
WINDOW
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Figure 1.3–1 A graphics window showing a plot.
OVERPLAY PLOT
the plot can be printed by selecting Print from the File menu on the graphics
window. The window can be closed by selecting Close on the File menu in the
graphics window. You will then be returned to the prompt in the Command
window.
Other useful plotting functions are title and gtext. These functions
place text on the plot. Both accept text within parentheses and single quotes, as
with the xlabel function. The title function places the text at the top of the
plot; the gtext function places the text at the point on the plot where the cursor
is located when you click the left mouse button.
You can create multiple plots, called overlay plots, by including another set
or sets of values in the plot function. For example, to plot the functions
y = 21x and z 4 sin 3x for 0 x 5 on the same plot, the session is
>>x = 0:0.01:5;
>>y = 2*sqrt(x);
>>z = 4*sin(3*x);
>>plot(x,y,x,z),xlabel(‘x’),gtext(‘y’),gtext(‘z’)
After the plot appears on the screen, the program waits for you to position
the cursor and click the mouse button, once for each gtext function used.
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Use the gtext function to place the labels y and z next to the appropriate
curves.
You can also distinguish curves from one another by using different line types
for each curve. For example, to plot the z curve using a dashed line, replace the
plot(x,y,x,z) function in the above session with plot(x,y,x,z, ‘ ’).
Other line types can be used. These are discussed in Chapter 5.
Sometimes it is useful or necessary to obtain the coordinates of a point on a
plotted curve. The function ginput can be used for this purpose. Place it at the
end of all the plot and plot formatting statements, so that the plot will be in its nal
form. The command [x,y] = ginput(n) gets n points and returns the x and
y coordinates in the vectors x and y, which have a length n. Position the cursor
using a mouse, and press the mouse button. The returned coordinates have the
same scale as the coordinates on the plot.
In cases where you are plotting data, as opposed to functions, you should use
data markers to plot each data point (unless there are very many data points). To
mark each point with a plus sign , the required syntax for the plot function is
plot(x,y,’’). You can connect the data points with lines if you wish. In
that case, you must plot the data twice, once with a data marker and once without
a marker.
For example, suppose the data for the independent variable is x =
[15:2:23] and the dependent variable values are y = [20, 50, 60, 90,
70]. To plot the data with plus signs, use the following session:
>>x = 15:2:23;
>>y = [20, 50, 60, 90, 70];
>>plot(x,y,’+’,x,y),xlabel(‘x’),ylabel(‘y’), grid
The grid command puts grid lines on the plot. Other data markers are available.
These are discussed in Chapter 5.
Table 1.3–3 summarizes these plotting commands. We will discuss other
plotting functions, and the Plot Editor, in Chapter 5.
Table 1.3–3 Some MATLAB plotting commands
Command
Description
[x,y] ginput(n)
Enables the mouse to get n points from a plot, and returns
the x and y coordinates in the vectors x and y, which have
a length n.
Puts grid lines on the plot.
Enables placement of text with the mouse.
Generates a plot of the array y versus the array x on
rectilinear axes.
Puts text in a title at the top of the plot.
Adds a text label to the horizontal axis (the abscissa).
Adds a text label to the vertical axis (the ordinate).
grid
gtext(‘text’)
plot(x,y)
title(‘text’)
xlabel(‘text’)
ylabel(‘text’)
25
DATA MARKER
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Test Your Understanding
T1.3–3 Use MATLAB to plot the function s = 2 sin(3t + 2) + 15t + 1 over
the interval 0 t 5. Put a title on the plot, and properly label the axes.
The variable s represents speed in feet per second; the variable t represents time in seconds.
T1.3–4 Use MATLAB to plot the functions y = 4 16x + 1 and z 5e0.3x 2x
over the interval 0 x 1.5. Properly label the plot and each curve. The
variables y and z represent force in newtons; the variable x represents
distance in meters.
Linear Algebraic Equations
You can use the left division operator (\) in MATLAB to solve sets of linear
algebraic equations. For example, consider the set
6x + 12y + 4z = 70
7x - 2y + 3z = 5
2x + 8y - 9z = 64
To solve such sets in MATLAB, you must create two arrays; we will call them
A and B. The array A has as many rows as there are equations and as many
columns as there are variables. The rows of A must contain the coef cients of x,
y, and z in that order. In this example, the rst row of A must be 6, 12, 4; the second row must be 7, 2, 3; and the third row must be 2, 8, 9. The array B contains the constants on the right-hand side of the equation; it has one column and
as many rows as there are equations. In this example, the rst row of B is 70, the
second is 5, and the third is 64. The solution is obtained by typing A\B. The
session is
>>A = [6,12,4;7,-2,3;2,8,-9];
>>B = [70;5;64];
>>Solution = A\B
Solution =
3
5
-2
The solution is x = 3, y = 5, and z = - 2.
This method works ne when the equation set has a unique solution. To learn
how to deal with problems having a nonunique solution (or perhaps no solution
at all!), see Chapter 8.
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Test Your Understanding
T1.3–5 Use MATLAB to solve the following set of equations.
6x - 4y + 8z = 112
- 5x - 3y + 7z = 75
14x + 9y - 5z = - 67
(Answer: x = 2, y = - 5, z = 10.)
1.4 Script Files and the Editor/Debugger
You can perform operations in MATLAB in two ways:
1. In the interactive mode, in which all commands are entered directly in the
Command window.
2. By running a MATLAB program stored in script le. This type of le
contains MATLAB commands, so running it is equivalent to typing all the
commands, one at a time, at the Command window prompt. You can run
the le by typing its name at the Command window prompt.
When the problem to be solved requires many commands or a repeated set of
commands, or has arrays with many elements, the interactive mode is inconvenient. Fortunately, MATLAB allows you to write your own programs to avoid
this dif culty . You write and save MATLAB programs in M- les, which have the
extension .m; for example, program1.m.
MATLAB uses two types of M- les: script les and function les. You can use
the Editor/Debugger built into MATLAB to create M- les. Because they contain
commands, script les are sometimes called command les. Function les are
discussed in Chapter 3.
SCRIPT FILE
Creating and Using a Script File
The symbol % designates a comment, which is not executed by MATLAB. Comments are used mainly in script les for the purpose of documenting the le. The
comment symbol may be put anywhere in the line. MATLAB ignores everything
to the right of the % symbol. For example, consider the following session.
>>% This is a comment.
>>x = 2+3 % So is this.
x =
5
Note that the portion of the line before the % sign is executed to compute x.
Here is a simple example that illustrates how to create, save, and run a script
le, using the Editor/Debugger built into MATLAB. However, you may use another
COMMENT
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Figure 1.4–1 The MATLAB Command window with the Editor/Debugger open.
text editor to create the le. The sample le is shown below. It computes the cosine
of the square root of several numbers and displays the results on the screen.
% Program Example_1.m
% This program computes the cosine of
% the square root and displays the result.
x = sqrt(13:3:25);
y = cos(x)
To create this new M- le in the Command window , select New from the File
menu, then select Blank M- le. You will then see a new edit window. This is the
Editor/Debugger window as shown in Figure 1.4–1. Type in the le as shown
above. You can use the keyboard and the Edit menu in the Editor/Debugger as
you would in most word processors to create and edit the le. When nished,
select Save from the File menu in the Editor/Debugger. In the dialog box that
appears, replace the default name provided (usually named Untitled) with the
name Example_1, and click on Save. The Editor/Debugger will automatically
provide the extension .m and save the le in the MATLAB current directory,
which for now we will assume to be on the hard drive.
Once the le has been saved, in the MA TLAB Command window type the
script le’ s name Example_1 to execute the program. You should see the result
displayed in the Command window. Figure 1.4–1 shows a screen containing the
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29
resulting Command window display and the Editor/Debugger opened to display
the script le.
Effective Use of Script Files
Create script les to avoid the need to retype lengthy and commonly used procedures. Here are some other things to keep in mind when using script les:
1. The name of a script le must follow the MA TLAB convention for naming
variables.
2. Recall that typing a variable’s name at the Command window prompt causes
MATLAB to display the value of that variable. Thus, do not give a script le
the same name as a variable it computes because MATLAB will not be able
to execute that script le more than once, unless you clear the variable.
3. Do not give a script le the same name as a MA TLAB command or function.
You can check to see if a command, function, or le name already exists by
using the exist command. For example, to see if a variable example1
already exists, type exist(‘example1’); this will return a 0 if the
variable does not exist and a 1 if it does. To see if an M- le example1.m
already exists, type exist(‘example1.m’,’ le’) before creating the
le; this will return a 0 if the le does not exist and a 2 if it does. Finally ,
to see if a built-in function example1 already exists, type exist
(‘example1’, ‘builtin’) before creating the le; this will return
a 0 if the built-in function does not exist and a 5 if it does.
Note that not all functions supplied with MATLAB are built-in functions.
For example, the function mean.m is supplied but is not a built-in function. The
command exist(‘mean.m’, ‘ le’) will return a 2, but the command
exist(‘mean’, ‘builtin’) will return a 0. You may think of built-in
functions as primitives that form the basis for other MATLAB functions. You cannot view the entire le of a built-in function in a text editor , only the comments.
Debugging Script Files
Debugging a program is the process of nding and removing the “bugs,” or errors,
in a program. Such errors usually fall into one of the following categories.
1. Syntax errors such as omitting a parenthesis or comma, or spelling a command name incorrectly. MATLAB usually detects the more obvious errors
and displays a message describing the error and its location.
2. Errors due to an incorrect mathematical procedure, called runtime errors.
They do not necessarily occur every time the program is executed; their
occurrence often depends on the particular input data. A common example
is division by zero.
To locate an error, try the following:
1. Always test your program with a simple version of the problem, whose
answers can be checked by hand calculations.
DEBUGGING
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2. Display any intermediate calculations by removing semicolons at the end
of statements.
3. Use the debugging features of the Editor/Debugger, which are introduced in
Chapter 4. However, one advantage of MATLAB is that it requires relatively
simple programs to accomplish many types of tasks. Thus you probably will
not need to use the Debugger for the problems encountered in this text.
Programming Style
Comments may be put anywhere in the script le. However , because the rst
comment line before any executable statement is the line searched by the
lookfor command, discussed later in this chapter, consider putting keywords that describe the script le in this rst line (called the H1 line). A suggested structure for a script le is the following.
1. Comments section In this section put comment statements to give
a. The name of the program and any keywords in the rst line.
b. The date created and the creators’ names in the second line.
c. The de nitions of the variable names for every input and output
variable. Divide this section into at least two subsections, one for input
data and one for output data. A third, optional section may include
de nitions of variables used in the calculations. Be sure to include the
units of measurement for all input and all output variables!
d. The name of every user-de ned function called by the program.
2. Input section In this section put the input data and/or the input functions
that enable data to be entered. Include comments where appropriate for
documentation.
3. Calculation section Put the calculations in this section. Include comments
where appropriate for documentation.
4. Output section In this section put the functions necessary to deliver the
output in whatever form required. For example, this section might contain
functions for displaying the output on the screen. Include comments where
appropriate for documentation.
The programs in this text often omit some of these elements to save space. Here the
text discussion associated with the program provides the required documentation.
Controlling Input and Output
MATLAB provides several useful commands for obtaining input from the user
and for formatting the output (the results obtained by executing the MATLAB
commands). Table 1.4–1 summarizes these commands.
The disp function (short for “display”) can be used to display the value of a
variable but not its name. Its syntax is disp(A), where A represents a MATLAB
variable name. The disp function can also display text such as a message to the
user. You enclose the text within single quotes. For example, the command
disp(‘The predicted speed is:’) causes the message to appear on
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Script Files and the Editor/Debugger
31
Table 1.4–1 Input/output commands
Command
Description
disp(A)
Displays the contents, but not the name, of the
array A.
Displays the text string enclosed within
single quotes.
Controls the screen’s output display format (see
Table 1.1–5).
Displays the text in quotes, waits for user input
from the keyboard, and stores the value in x.
Displays the text in quotes, waits for user input from
the keyboard, and stores the input as a string in x.
Displays a menu whose title is in the string
variable ‘title’ and whose choices are
‘option1’,‘option2’, and so on.
disp(‘text’)
format
x = input(‘text’)
x = input(‘text’,’s’)
k=menu(‘title’,’option1’,
’option2’,...
the screen. This command can be used with the rst form of the disp function
in a script le as follows (assuming the value of Speed is 63):
disp(‘The predicted speed is:’)
disp(Speed)
When the le is run, these lines produce the following on the screen:
The predicted speed is:
63
The input function displays text on the screen, waits for the user to enter
something from the keyboard, and then stores the input in the speci ed variable.
For example, the command x = input(‘Please enter the value of
x:’) causes the message to appear on the screen. If you type 5 and press Enter,
the variable x will have the value 5.
A string variable is composed of text (alphanumeric characters). If you want
to store a text input as a string variable, use the other form of the input command.
For example, the command Calendar = input(‘Enter the day of
the week:’,’s’) prompts you to enter the day of the week. If you type
Wednesday, this text will be stored in the string variable Calendar.
Use the menu function to generate a menu of choices for user input. Its
syntax is
k = menu(‘title’,’option1’,’option2’,...)
The function displays the menu whose title is in the string variable ‘title’ and
whose choices are string variables ‘option1’, ‘option2’, and so on. The
returned value of k is 1, 2, . . . depending on whether you click on the button for
option1, option2, and so forth. For example, the following script uses a menu
to select the data marker for a graph, assuming that the arrays x and y already exist.
k = menu(‘Choose a data marker’,’o’,’*’,’x’);
type = [‘o’,’*’,’x’];
plot(x,y,x,y,type(k))
STRING VARIABLE
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Test Your Understanding
T1.4–1 The surface area A of a sphere depends on its radius r as follows:
A ⫽ 4␲r2. Write a script le that prompts the user to enter a radius, computes the surface area, and displays the result.
Example of a Script File
The following is a simple example of a script le that shows the preferred program style. The speed ␷ of a falling object dropped with no initial velocity is given
as a function of time t by ␷ = gt, where g is the acceleration due to gravity. In
SI units, g = 9.81 m/s2. We want to compute and plot ␷ as a function of t for
0 … t … tfinal, where t nal is the nal time entered by the user . The script le is
the following.
% Program Falling_Speed.m: plots speed of a falling object.
% Created on March 1, 2009 by W. Palm III
%
% Input Variable:
% t nal = nal time (in seconds)
%
% Output Variables:
% t = array of times at which speed is computed (seconds)
% v = array of speeds (meters/second)
%
% Parameter Value:
g = 9.81; % Acceleration in SI units
%
% Input section:
tfinal = input(‘Enter the nal time in seconds:’);
%
% Calculation section:
dt = tfinal/500;
t = 0:dt:tfinal; % Creates an array of 501 time values.
v = g*t;
%
% Output section:
plot(t,v),xlabel(‘Time (seconds)’),ylabel(‘Speed (meters/second)’)
After creating this le, you save it with the name Falling_Speed.m. To
run it, you type Falling_Speed (without the .m) in the Command window at
the prompt. You will then be asked to enter a value for t nal . After you enter a
value and press Enter, you will see the plot on the screen.
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1.5
The MATLAB Help System
1.5 The MATLAB Help System
To explore the more advanced features of MATLAB not covered in this book,
you will need to know how to use effectively the MATLAB Help System.
MATLAB has these options to get help for using MathWorks products.
1. Function Browser This provides quick access to the documentation for the
MATLAB function.
2. Help Browser This graphical user interface helps you nd information and
view online documentation for your MathWorks products.
3. Help Functions The functions help, lookfor, and doc can be used to
display syntax information for a speci ed function.
4. Other Resources For additional help, you can run demos, contact technical
support, search documentation for other MathWorks products, view a list
of other books, and participate in a newsgroup.
The Function Browser
To activate the Function Browser, either select Function Browser from the Help
menu or select the fx icon to the left of the prompt. Figure 1.5–1 shows the resulting menu after the Graphics category has been selected. The subwindow shown
opens when the plot function is selected. Scroll down to see the entire documentation of the plot function.
The Help Browser
To open the Help Browser, select Product Help from the Help menu, or click the
question mark button in the toolbar. The Help Browser contains two window
Figure 1.5–1 The Function Browser after plot has been selected.
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Figure 1.5–2 The MATLAB Help Browser.
“panes”: the Help Navigator pane on the left and the Display pane on the right
(see Figure 1.5–2). The Help Navigator contains two tabs:
■
■
Contents: a contents listing tab
Search Results: a search tab having a nd function and full text search
features
Use the tabs in the Help Navigator to nd documentation. You view documentation
in the Display pane. To open the Help Navigator pane from the display pane, click
on Help Navigator in the View menu.
Finding Documentation
Figure 1.5–3 shows the result of clicking on the sign next to MATLAB in the Help
Navigator. A submenu appears that shows the various Help topics for MATLAB.
Viewing Documentation
After nding documentation with the Help Navigator , view the documentation in
the Display pane. While viewing a page of documentation, you can
■
■
■
Scroll to see contents not currently visible in the window.
View the previous or next page in the document by clicking the left or right
arrow at the top of the page.
View the previous or next item in the index by clicking the left or right
arrow at the bottom of the page.
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Figure 1.5–3 The Help Navigator showing the submenus under the
MATLAB category.
■
Find a phrase or speci ed term by typing the term in the Search box
below the Help Browser toolbar and pressing the Enter key. The left-hand
pane will then display all the Help pages and function documentation that
contains the speci ed term.
Using the Contents Tab
Click the Contents tab in the Help Navigator to list the titles and table of contents for all product documentation. To expand the listing for an item, click the
to the left of the item. To collapse the listings for an item, click the to the
left of the item, or double-click the item. Click on an item to select it. The rst
page of that document appears in the Display pane. Double-clicking an item in
the contents listing expands the listing for that item and shows the rst page of
that document in the Display pane.
The Contents pane is synchronized with the Display pane. By default, the
item selected in the Contents pane always matches the documentation appearing
in the Display pane. Thus, the contents tree is synchronized with the displayed
document.
Using the Search Results Tab
Click the Search Results tab in the Help Navigator pane to nd all MA TLAB
documents containing a speci ed phrase. Type the phrase in the “Search” box.
Then press Enter. The list of documents and the heading under which the phrase
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is found in that document then appear in the Help Navigator pane. Select an entry
from the list of results to view that document in the Display pane.
Figure 1.5–4 shows the results of typing plot in the “Search” box. The
Display pane shows the documentation for the plot function (scroll down to
see all of it), and the Help Navigator pane shows the Documentation Search
results and the Demo Search results.
Help Functions
Three MATLAB functions can be used for accessing online information about
MATLAB functions.
The help Function The help function is the most basic way to determine
the syntax and behavior of a particular function. For example, typing help
log10 in the Command window produces the following display:
LOG10 Common (base 10) logarithm.
LOG10(X) is the base 10 logarithm of the elements of X.
Complex results are produced if X is not positive.
See also LOG, LOG2, EXP, LOGM.
Note that the display describes what the function does, warns about any unexpected
results if nonstandard argument values are used, and directs the user to other related
functions.
All the MATLAB functions are organized into logical groups, upon which the
MATLAB directory structure is based. For instance, all elementary mathematical
Figure 1.5–4 The results of entering plot in the “Search” box.
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The MATLAB Help System
functions such as log10 reside in the elfun directory, and the polynomial
functions reside in the polyfun directory. To list the names of all the functions
in that directory, with a brief description of each, type help polyfun. If you
are unsure of what directory to search, type help to obtain a list of all the directories, with a description of the function category each represents.
Typing helpwin topic displays the Help text for the speci ed topic
inside the Desktop Help Browser window. Links are created to functions referenced in the “See Also” line of the Help text. You can also access the Help window by selecting the Help option under the Help menu, or by clicking the
question mark button on the toolbar.
The lookfor Function
The lookfor function allows you to search for
functions on the basis of a keyword. It searches through the rst line of Help text,
known as the H1 line, for each MATLAB function, and returns all the H1 lines
containing a speci ed keyword. For example, MA TLAB does not have a function named sine. So the response from help sine is
sine.m not found
However, typing lookfor sine produces over a dozen matches, depending on
which toolboxes you have installed. For example, you will see, among others,
ACOS
ACOSD
ACOSH
ASIN
...
SIN
...
Inverse cosine, result in radians
Inverse cosine, result in degrees
Inverse hyperbolic cosine
Inverse sine, result in radians
Sine of argument in radians
From this list you can nd the correct name for the sine function. Note that all
words containing sine are returned, such as cosine. Adding -all to the lookfor
function searches the entire Help entry, not just the H1 line.
The doc Function Typing doc function displays the documentation for the
MATLAB function function. Typing doc toolbox/function displays
the documentation for the speci ed toolbox function. Typing doc toolbox
displays the documentation road map page for the speci ed toolbox.
The MathWorks Website
If your computer is connected to the Internet, you can access The MathWorks,
Inc., the home of MATLAB. You can use electronic mail to ask questions, make
suggestions, and report possible bugs. You can also use a solution search engine
at The MathWorks website to query an up-to-date database of technical support
information. The website address is http://www.mathworks.com.
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Table 1.5–1 MATLAB Help functions
Function
Use
doc
Displays the start page of the documentation in the Help
Browser.
Displays the documentation for the MATLAB function
function.
Displays the documentation for the speci ed toolbox
function.
Displays the documentation road map page for the speci ed
toolbox.
Displays a list of all the function directories, with a description of the function category each represents.
Displays in the Command window a description of the
speci ed function function.
Displays the Help text for the speci ed topic inside the
Desktop Help Browser window.
Displays in the Command window a brief description for
all functions whose description includes the speci ed
keyword topic.
Displays the M- le lename without opening it with a
text editor.
doc function
doc toolbox/
function
doc toolbox
help
help function
helpwin topic
lookfor topic
type
lename
The Help system is very powerful and detailed, so we have only described
its basics. You can, and should, use the Help system to learn how to use its features in greater detail.
Table 1.5–1 summarizes the MATLAB Help functions.
1.6 Problem-Solving Methodologies
Designing new engineering devices and systems requires a variety of problemsolving skills. (This variety is what keeps engineering from becoming boring!)
When you are solving a problem, it is important to plan your actions ahead of
time. You can waste many hours by plunging into the problem without a plan of
attack. Here we present a plan of attack, or methodology, for solving engineering
problems in general. Because solving engineering problems often requires a
computer solution and because the examples and exercises in this text require
you to develop a computer solution (using MATLAB), we also discuss a methodology for solving computer problems in particular.
Steps in Engineering Problem Solving
MODEL
Table 1.6–1 summarizes the methodology that has been tried and tested by the
engineering profession for many years. These steps describe a general problemsolving procedure. Simplifying the problem suf ciently and applying the appropriate fundamental principles is called modeling, and the resulting mathematical
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Table 1.6–1 Steps in engineering problem solving
1. Understand the purpose of the problem.
2. Collect the known information. Realize that some of it might later be found
unnecessary.
3. Determine what information you must nd.
4. Simplify the problem only enough to obtain the required information. State any assumptions you make.
5. Draw a sketch and label any necessary variables.
6. Determine which fundamental principles are applicable.
7. Think generally about your proposed solution approach and consider other approaches
before proceeding with the details.
8. Label each step in the solution process.
9. If you solve the problem with a program, hand check the results using a simple version
of the problem. Checking the dimensions and units and printing the results of intermediate steps in the calculation sequence can uncover mistakes.
10. Perform a “reality check” on your answer. Does it make sense? Estimate the range of
the expected result and compare it with your answer. Do not state the answer with
greater precision than is justi ed by any of the following:
(a) The precision of the given information.
(b) The simplifying assumptions.
(c) The requirements of the problem.
Interpret the mathematics. If the mathematics produces multiple answers, do not discard
some of them without considering what they mean. The mathematics might be trying to
tell you something, and you might miss an opportunity to discover more about the
problem.
description is called a mathematical model, or just a model. When the modeling
is nished, we need to solve the mathematical model to obtain the required answer. If the model is highly detailed, we might need to solve it with a computer
program. Most of the examples and exercises in this text require you to develop
a computer solution (using MATLAB) to problems for which the model has
already been developed. Thus we will not always need to use all the steps shown
in Table 1.6–1. Greater discussion of engineering problem solving can be found
in [Eide, 2008].*
Example of Problem Solving
Consider the following simple example of the steps involved in problem solving.
Suppose you work for a company that produces packaging. You are told that a
new packaging material can protect a package when dropped, provided that the
package hits the ground at less than 25 ft/sec. The package’s total weight is 20 lb,
and it is rectangular with dimensions of 12 by 12 by 8 in. You must determine
whether the packaging material provides enough protection when the package is
carried by delivery persons.
*
References appear in Appendix D.
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The steps in the solution are as follows:
1. Understand the purpose of the problem. The implication here is that the
packaging is intended to protect against being dropped while the delivery
person is carrying it. It is not intended to protect against the package falling
off a moving delivery truck. In practice, you should make sure that the person giving you this assignment is making the same assumption. Poor communication is the cause of many errors!
2. Collect the known information. The known information is the package’s
weight, dimensions, and maximum allowable impact speed.
3. Determine what information you must nd. Although it is not explicitly
stated, you need to determine the maximum height from which the package
can be dropped without damage. You need to nd a relationship between
the speed of impact and the height at which the package is dropped.
4. Simplify the problem only enough to obtain the required information.
State any assumptions you make. The following assumptions will
simplify the problem and are consistent with the problem statement as we
understand it:
a. The package is dropped from rest with no vertical or horizontal velocity.
b. The package does not tumble (as it might when dropped from a moving
truck). The given dimensions indicate that the package is not thin and
thus will not “ utter” as it falls.
c. The effect of air drag is negligible.
d. The greatest height from which the delivery person could drop the
package is 6 ft (and thus we ignore the existence of a delivery person
8 ft tall!).
e. The acceleration g due to gravity is constant (because the distance
dropped is only 6 ft).
5. Draw a sketch and label any necessary variables. Figure 1.6–1 is a sketch
of the situation, showing the height h of the package, its mass m, its speed
␷, and the acceleration due to gravity g.
6. Determine which fundamental principles are applicable. Because this
problem involves a mass in motion, we can apply Newton’s laws. From
physics we know that the following relations result from Newton’s laws
and the basic kinematics of an object falling a short distance under the inuence of gravity , with no air drag or initial velocity:
a. Height versus time to impact ti: h = 12gt2i .
b. Impact speed ␷i versus time to impact: ␷i = gti.
c. Conservation of mechanical energy: mgh = 12 m␷i2.
7. Think generally about your proposed solution approach and consider other
approaches before proceeding with the details. We could solve the second
equation for ti and substitute the result into the rst equation to obtain the
relation between h and ␷i. This approach would also allow us to nd the
time to drop ti. However, this method involves more work than necessary
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Problem-Solving Methodologies
Package
m
v
h
g
Ground
Figure 1.6–1 Sketch of the
dropped-package problem.
because we need not nd the value of ti. The most ef cient approach is to
solve the third relation for h.
h =
1 ␷i2
2 g
(1.6–1)
Notice that the mass m cancels out of the equation. The mathematics just
told us something! It told us that the mass does not affect the relation
between the impact speed and the height dropped. Thus we do not need the
weight of the package to solve the problem.
8. Label each step in the solution process. This problem is so simple that
there are only a few steps to label:
a. Basic principle: conservation of mechanical energy
1 ␷i2
h =
2 g
b. Determine the value of the constant g: g = 32.2 ft/sec2.
c. Use the given information to perform the calculation and round off the
result consistent with the precision of the given information:
h =
1 252
= 9.7 ft
2 32.2
Because this text is about MATLAB, we might as well use it to do this
simple calculation. The session looks like this:
>>g=32.2;
>>vi=25;
>>h=vi^2/(2*g)
h =
9.7050
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9. Check the dimensions and units. This check proceeds as follows, using
Equation (1.6–1),
[ft]2 [sec]2
1 [ft/sec]2
=
= [ft]
[ft] = c d
2 [ft/sec2]
[sec]2 [ft]
which is correct.
10. Perform a reality check and precision check on the answer. If the
computed height were negative, we would know that we did something
wrong. If it were very large, we might be suspicious. However, the
computed height of 9.7 ft does not seem unreasonable.
If we had used a more accurate value for g, say g 32.17, then we
would be justi ed in rounding the result to h 9.71. However, given the
need to be conservative here, we probably should round the answer down
to the nearest foot. So we probably should report that the package will not
be damaged if it is dropped from a height of less than 9 ft.
The mathematics told us that the package mass does not affect the
answer. The mathematics did not produce multiple answers here. However,
many problems involve the solution of polynomials with more than one
root; in such cases we must carefully examine the signi cance of each.
Steps for Obtaining a Computer Solution
If you use a program such as MATLAB to solve a problem, follow the steps
shown in Table 1.6–2. Greater discussion of modeling and computer solutions
can be found in [Star eld, 1990] and [Jayaraman, 1991].
MATLAB is useful for doing numerous complicated calculations and then
automatically generating a plot of the results. The following example illustrates
the procedure for developing and testing such a program.
Table 1.6–2 Steps for developing a computer solution
1. State the problem concisely.
2. Specify the data to be used by the program. This is the input.
3. Specify the information to be generated by the program. This is the output.
4. Work through the solution steps by hand or with a calculator; use a simpler set of data if
necessary.
5. Write and run the program.
6. Check the output of the program with your hand solution.
7. Run the program with your input data and perform a “reality check” on the output. Does
it make sense? Estimate the range of the expected result and compare it with your
answer.
8. If you will use the program as a general tool in the future, test it by running it for a
range of reasonable data values; perform a reality check on the results.
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Problem-Solving Methodologies
Piston Motion
43
EXAMPLE 1.6–1
Figure 1.6–2a shows a piston, connecting rod, and crank for an internal combustion engine. When combustion occurs, it pushes the piston down. This motion causes the connecting rod to turn the crank, which causes the crankshaft to rotate. We want to develop a
MATLAB program to compute and plot the distance d traveled by the piston as a function
of the angle A, for given values of lengths L1 and L2. Such a plot would help the engineers
designing the engine to select appropriate values for lengths L1 and L2.
We are told that typical values for these lengths are L1 1 ft and L2 0.5 ft. Because
the mechanism’s motion is symmetrical about A 0, we need consider only angles in the
range 0 A 180°. Figure 1.6–2b shows the geometry of the motion. From this gure
we can use trigonometry to write the following expression for d:
(1.6–2)
d = L1 cos B + L2 cos A
Thus to compute d given the lengths L1 and L2 and the angle A, we must rst determine
the angle B. We can do so using the law of sines, as follows:
sin A
sin B
=
L1
L2
Solve this for B:
sin B =
L2 sin A
L1
B = sin - 1 a
L2 sin A
b
L1
(1.6–3)
Piston
L1
L1
Connecting
Rod
B
B
A
d
A
L2
Crank
L2
Crankshaft
(a)
(b)
Figure 1.6–2 A piston, connecting rod, and crank for an
internal combustion engine.
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Equations (1.6–2) and (1.6–3) form the basis of our calculations. Develop and test a
MATLAB program to plot d versus A.
■ Solution
Here are the steps in the solution, following those listed in Table 1.6–2.
1.
2.
3.
4.
5.
State the problem concisely. Use Equations (1.6–2) and (1.6–3) to compute d; use
enough values of A in the range 0 A 180° to generate an adequate (smooth) plot.
Specify the input data to be used by the program. The lengths L1 and L2 and the
angle A are given.
Specify the output to be generated by the program. A plot of d versus A is the
required output.
Work through the solution steps by hand or with a calculator. You could have
made an error in deriving the trigonometric formulas, so you should check them for
several cases. You can check for these errors by using a ruler and protractor to
make a scale drawing of the triangle for several values of the angle A; measure the
length d; and compare it to the calculated values. Then you can use these results to
check the output of the program.
Which values of A should you use for the checks? Because the triangle
“collapses” when A 0° and A 180°, you should check these cases. The results
are d L1 L2 for A 0° and d L1 L2 for A 180°. The case A 90° is
also easily checked by hand, using the Pythagorean theorem; for this case
d = 2L21 - L22. You should also check one angle in the quadrant 0° A 90°
and one in the quadrant 90° A 180°. The following table shows the results of
these calculations using the given typical values: L1 1, L2 0.5 ft.
A (degrees)
d (ft)
0
60
90
120
180
1.5
1.15
0.87
0.65
0.5
Write and run the program. The following MATLAB session uses the values
L1 1, L2 0.5 ft.
>>L_1 = 1;
>>L_2 = 0.5;
>>R = L_2/L_1;
>>A_d = 0:0.5:180;
>>A_r = A_d*(pi/180);
>>B = asin(R*sin(A_r));
>>d = L_1*cos(B)+L_2*cos(A_r);
>>plot(A_d,d),xlabel(‘A (degrees)’), ...
ylabel(‘d (feet)’),grid
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1.5
1.4
1.3
1.2
d (feet)
1.1
1
0.9
0.8
0.7
0.6
0.5
0
20
40
60
80
100
A (degrees)
120
140
160
180
Figure 1.6–3 Plot of the piston motion versus crank angle.
Note the use of the underscore (_) in variable names to make the names more
meaningful. The variable A_d represents the angle A in degrees. Line 4 creates an
array of numbers 0, 0.5, 1, 1.5, Á ,180. Line 5 converts these degree values to
radians and assigns the values to the variable A_r. This conversion is necessary
because MATLAB trigonometric functions use radians, not degrees. (A common
oversight is to use degrees.) MATLAB provides the built-in constant pi to use
for . Line 6 uses the inverse sine function asin.
The plot command requires the label and grid commands to be on the same
line, separated by commas. The line-continuation operator, called an ellipsis,
consists of three periods. This operator enables you to continue typing the line after
you press Enter. Otherwise, if you continued typing without using the ellipsis, you
would not see the entire line on the screen. Note that the prompt is not visible when
you press Enter after the ellipsis.
The grid command puts grid lines on the plot so that you can read values
from the plot more easily. The resulting plot appears in Figure 1.6–3.
6.
Check the output of the program with your hand solution. Read the values from
the plot corresponding to the values of A given in the preceding table. You can use
the ginput function to read values from the plot. The values should agree with
one another, and they do.
7.
Run the program and perform a reality check on the output. You might suspect an
error if the plot showed abrupt changes or discontinuities. However, the plot is
smooth and shows that d behaves as expected. It decreases smoothly from its maximum at A 0° to its minimum at A 180°.
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8.
Test the program for a range of reasonable input values. Test the program using
various values for L1 and L2, and examine the resulting plots to see whether they
are reasonable. Something you might try on your own is to see what happens if
L1 L2. Should the mechanism work the same way it does when L1 L2? What
does your intuition tell you to expect from the mechanism? What does the program
predict?
1.7 Summary
You should now be familiar with basic operations in MATLAB. These include
■
■
■
Starting and exiting MATLAB
Computing simple mathematical expressions
Managing variables
You should also be familiar with the MATLAB menu and toolbar system.
The chapter gives an overview of the various types of problems MATLAB
can solve. These include
■ Using arrays and polynomials
■ Creating plots
■ Creating script les
Table 1.7–1 is a guide to the tables of this chapter. The following chapters give
more details on these topics.
Table 1.7–1 Guide to commands and features introduced in this chapter
Scalar arithmetic operations
Order of precedence
Commands for managing the work session
Special variables and constants
Numeric display formats
Some commonly used mathematical functions
System, directory, and le commands
Some MATLAB plotting commands
Input/output commands
MATLAB Help functions
Table 1.1–1
Table 1.1–2
Table 1.1–3
Table 1.1–4
Table 1.1–5
Table 1.3–1
Table 1.3–2
Table 1.3–3
Table 1.4–1
Table 1.5–1
Key Terms with Page References
Argument, 8
Array, 19
Array index, 19
ASCII les, 21
Command window, 6
Comment, 27
Current directory, 16
Data le, 21
Data marker, 25
Debugging, 29
Desktop, 5
Graphics window, 23
MAT- les, 20
Model, 39
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Problems
Search path, 22
Session, 7
String variable, 31
Variable, 7
Workspace, 11
Overlay plot, 24
Path, 22
Precedence, 9
Scalar, 8
Script le, 27
Problems
Answers to problems marked with an asterisk are given at the end of the text.
Section 1.1
1.
Make sure you know how to start and quit a MATLAB session. Use
MATLAB to make the following calculations, using the values x 10,
y 3. Check the results by using a calculator.
a. u x y
b. v xy
c. w x y
d. z sin x
e. r 8 sin y
f. s 5 sin (2y)
2.*
Suppose that x 2 and y 5. Use MATLAB to compute the following.
yx3
3
x5
3x
a.
b.
c. xy
d. 5
x - y
2y
2
x - 1
3.
Suppose that x 3 and y 4. Use MATLAB to compute the following,
and check the results with a calculator.
3y
4(y - 5)
1 -1
a. a1 - 5 b
b. 3x2
c.
d.
4x - 8
3x - 6
x
4.
Evaluate the following expressions in MATLAB for the given value of x.
Check your answers by hand.
a. y = 6x3 +
4
,
x
x3
(4x)2
,
x9
25
e. y 7(x1/3) 4x 0.58,
b. y =
d. y = 2
c. y =
5.
x
3,
4
sin x
,
5
x7
x4
x 30
Assuming that the variables a, b, c, d, and f are scalars, write MATLAB
statements to compute and display the following expressions. Test your
statements for the values a 1.12, b 2.34, c 0.72, d 0.81, and
f 19.83.
b - a
a
c
s =
x = 1 + + 2
b
d - c
f
1
1 f2
r = 1
y
=
ab
c 2
+ 1 + 1 + 1
a
b
c
d
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6.
Use MATLAB to calculate
a.
3
45
(6) (72) + 3
4
7 - 145
c.
272
3194>5
+
+ 60(14)-3
4
5
b.
48.2(55) - 93
53 + 142
Check your answers with a calculator.
7.
The volume of a sphere is given by V 4r3/3, where r is the radius. Use
MATLAB to compute the radius of a sphere having a volume 40 percent
greater than that of a sphere of radius 4 ft.
8.*
Suppose that x 7 5i and y 4 3i. Use MATLAB to compute
a. x y
b. xy
c. x/ y
Use MATLAB to compute the following. Check your answers by hand.
9.
5 + 4i
5 - 4i
3
d.
2i
a. (3 6i)(7 9i)
c.
b.
3
i
2
10.
Evaluate the following expressions in MATLAB, for the values x 5 8i, y 6 7i. Check your answers by hand.
a. u x y
b. v xy
c. w x / y
x
y
d. z e
e. r = 1
f. s xy2
11.
The ideal gas law provides one way to estimate the pressure exerted by a
gas in a container. The law is
nRT
P =
V
More accurate estimates can be made with the van der Waals equation
P =
an2
nRT
- 2
V - nb
V
where the term nb is a correction for the volume of the molecules and the
term an2/V 2 is a correction for molecular attractions. The values of a and
b depend on the type of gas. The gas constant is R, the absolute temperature is T, the gas volume is V, and the number of gas molecules is indicated by n. If n 1 mol of an ideal gas were con ned to a volume of
V 22.41 L at 0°C (273.2 K), it would exert a pressure of 1 atm. In these
units, R 0.08206.
For chlorine (Cl2 ), a 6.49 and b 0.0562. Compare the pressure estimates given by the ideal gas law and the van der Waals equation for 1 mol
of Cl2 in 22.41 L at 273.2 K. What is the main cause of the difference in the
two pressure estimates, the molecular volume or the molecular attractions?
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Problems
12.
The ideal gas law relates the pressure P, volume V, absolute temperature
T, and amount of gas n. The law is
nRT
P =
V
where R is the gas constant.
An engineer must design a large natural gas storage tank to be
expandable to maintain the pressure constant at 2.2 atm. In December
when the temperature is 4°F (15°C), the volume of gas in the tank is
28 500 ft3. What will the volume of the same quantity of gas be in July
when the temperature is 88°F (31°C)? (Hint: Use the fact that n, R, and P
are constant in this problem. Note also that K °C 273.2.)
Section 1.3
13.
14.
15.
Suppose x takes on the values x 1, 1.2, 1.4, . . . , 5. Use MATLAB to
compute the array y that results from the function y 7 sin(4x). Use
MATLAB to determine how many elements are in the array y and the
value of the third element in the array y.
Use MATLAB to determine how many elements are in the array
sin(-pi/2):0.05:cos(0). Use MATLAB to determine the
10th element.
Use MATLAB to calculate
3
4
4
a. e(-2.1) + 3.47 log(14) + 1287
b. (3.4)7 log(14) + 1287
4.12
4.12 2
b
b
c. cos2 a
d. cos a
6
6
Check your answers with a calculator.
16.
Use MATLAB to calculate
b. 5 tan [3 sin1(13/5)]
a. 6 tan1(12.5) 4
c. 5 ln(7)
d. 5 log(7)
Check your answers with a calculator.
17.
The Richter scale is a measure of the intensity of an earthquake. The
energy E (in joules) released by the quake is related to the magnitude M
on the Richter scale as follows.
E = 104.4101.5M
How much more energy is released by a magnitude 7.6 quake than a 5.6
quake?
18.* Use MATLAB to nd the roots of 13 x3 182x2 184x 2503 0.
19.
Use MATLAB to nd the roots of the polynomial 70 x3 24x2 10x 20.
20.
Determine which search path MATLAB uses on your computer. If you use
a lab computer as well as a home computer, compare the two search paths.
Where will MATLAB look for a user-created M- le on each computer?
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21.
Use MATLAB to plot the function T ⫽ 6 ln t ⫺ 7e0.2t over the interval 1 ⱕ
t ⱕ 3. Put a title on the plot and properly label the axes. The variable T
represents temperature in degrees Celsius; the variable t represents time in
minutes.
22.
Use MATLAB to plot the functions u ⫽ 2 log10(60x ⫹ 1) and ␷ ⫽ 3 cos(6x)
over the interval 0 ⱕ x ⱕ 2. Properly label the plot and each curve. The variables u and ␷ represent speed in miles per hour; the variable x represents
distance in miles.
23.
The Fourier series is a series representation of a periodic function in terms
of sines and cosines. The Fourier series representation of the function
f (x) = e
is
1
0<x<␲
-1 -␲ < x < 0
4 sin x
sin 3x
sin 5x
sin 7x
+
+
+
+ Áb
a
␲
1
3
5
7
Plot on the same graph the function f (x) and its series representation,
using the four terms shown.
24.
A cycloid is the curve described by a point P on the circumference of a
circular wheel of radius r rolling along the x axis. The curve is described
in parametric form by the equations
x = r (␾ - sin ␾)
y = r (1 - cos ␾)
Use these equations to plot the cycloid for r ⫽ 10 in. and 0 ⱕ ␾ ⱕ 4␲.
Section 1.4
25.
A fence around a eld is shaped as shown in Figure P25. It consists of a
rectangle of length L and width W and a right triangle that is symmetric
about the central horizontal axis of the rectangle. Suppose the width W is
known (in meters) and the enclosed area A is known (in square meters).
Write a MATLAB script le in terms of the given variables W and A to
determine the length L required so that the enclosed area is A. Also
determine the total length of fence required. Test your script for the values
W ⫽ 6 m and A ⫽ 80 m2.
L
D
W
Figure P25
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Problems
26.
The four-sided gure shown in Figure P26 consists of two triangles having a common side a. The law of cosines for the top triangle states that
a2 = b21 + c21 - 2b1c1 cos A1
and a similar equation can be written for the bottom triangle. Develop a
procedure for computing the length of side c2 if you are given the lengths
of sides b1, b2, and c1 and the angles A1 and A2 in degrees. Write a script
le to implement this procedure. Test your script, using the following values: b1 180 m, b2 165 m, c1 115 m, A1 120°, and A2 100°.
b1
A1
C1
c1
a
C2
B1
b2
B2
A2
c2
Figure P26
Section 1.5
27.
Use the MATLAB Help facilities to nd information about the following
topics and symbols: plot, label, cos, cosine, :, and *.
28.
Use the MATLAB Help facilities to determine what happens if you use
the sqrt function with a negative argument.
29.
Use the MATLAB Help facilities to determine what happens if you use
the exp function with an imaginary argument.
Section 1.6
30.
31.
a. With what initial speed must you throw a ball vertically for it to reach
a height of 20 ft? The ball weighs 1 lb. How does your answer change
if the ball weighs 2 lb?
b. Suppose you want to throw a steel bar vertically to a height of 20 ft. The
bar weighs 2 lb. How much initial speed must the bar have to reach this
height? Discuss how the length of the bar affects your answer.
Consider the motion of the piston discussed in Example 1.6–1. The piston
stroke is the total distance moved by the piston as the crank angle varies
from 0° to 180°.
a. How does the piston stroke depend on L1 and L2?
b. Suppose L2 0.5 ft. Use MATLAB to plot the piston motion versus
crank angle for two cases: L1 0.6 ft and L1 1.4 ft. Compare each
plot with the plot shown in Figure 1.6–3. Discuss how the shape of
the plot depends on the value of L1.
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Photo courtesy of Stratosphere Corporation.
Engineering in the 21st Century. . .
Innovative Construction
W
e tend to remember the great civilizations of the past in part by their
public works, such as the Egyptian pyramids and the medieval cathedrals of Europe, which were technically challenging to create. Perhaps
it is in our nature to “push the limits,” and we admire others who do so. The challenge of innovative construction continues today. As space in our cities becomes
scarce, many urban planners prefer to build vertically rather than horizontally.
The newest tall buildings push the limits of our abilities, not only in structural
design but also in areas that we might not think of, such as elevator design and
operation, aerodynamics, and construction techniques. The photo above shows
the 1149-ft-high Las Vegas Stratosphere Tower, the tallest observation tower in
the United States. It required many innovative techniques in its assembly. The
construction crane shown in use is 400 ft tall.
Designers of buildings, bridges, and other structures will use new technologies and new materials, some based on nature’s designs. Pound for pound, spider
silk is stronger than steel, and structural engineers hope to use cables of synthetic
spider silk bers to build earthquake-resistant suspension bridges. Smart structures, which can detect impending failure from cracks and fatigue, are now close
to reality, as are active structures that incorporate powered devices to counteract
wind and other forces. The MATLAB Financial toolbox is useful for nancial
evaluation of large construction projects, and the MATLAB Partial Differential
Equation toolbox can be used for structural design. ■
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C H A P T E R
2
Numeric, Cell, and
Structure Arrays
OUTLINE
2.1 One- and Two-Dimensional Numeric Arrays
2.2 Multidimensional Numeric Arrays
2.3 Element-by-Element Operations
2.4 Matrix Operations
2.5 Polynomial Operations Using Arrays
2.6 Cell Arrays
2.7 Structure Arrays
2.8 Summary
Problems
One of the strengths of MATLAB is the capability to handle collections of items,
called arrays, as if they were a single entity. The array-handling feature means
that MATLAB programs can be very short.
The array is the basic building block in MATLAB. The following classes of
arrays are available in MATLAB 7:
numeric
character
logical
Array
cell structure
function handle
Java
So far we have used only numeric arrays, which are arrays containing only
numeric values. Within the numeric class are the subclasses single (single precision), double (double precision), int8, int16, and int32 (signed 8-bit, 16-bit, and
32-bit integers), and uint8, uint16, and uint32 (unsigned 8-bit, 16-bit, and 32-bit
integers). A character array is an array containing strings. The elements of logical
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arrays are “true” or “false,” which, although represented by the symbols 1 and
0, are not numeric quantities. We will study the logical arrays in Chapter 4. Cell
arrays and structure arrays are covered in Sections 2.6 and 2.7. Function handles
are treated in Chapter 3. The Java class is not covered in this text.
The rst four sections of this chapter treat concepts that are essential to understanding MATLAB and therefore must be covered. Section 2.5 treats polynomial
applications. Sections 2.6 and 2.7 introduce two types of arrays that are useful for
some specialized applications.
2.1 One- and Two-Dimensional Numeric Arrays
ROW VECTOR
COLUMN VECTOR
We can represent the location of a point in three-dimensional space by three
Cartesian coordinates x, y, and z. These three coordinates specify a vector p. (In
mathematical text we often use boldface type to indicate vectors.) The set of unit
vectors i, j, k, whose lengths are 1 and whose directions coincide with the x, y,
and z axes, respectively, can be used to express the vector mathematically as follows: p xi yj zk. The unit vectors enable us to associate the vector
components x, y, z with the proper coordinate axes; therefore, when we write p 5i 7j 2k, we know that the x, y, and z coordinates of the vector are 5, 7, and
2, respectively. We can also write the components in a speci c order , separate
them with a space, and identify the group with brackets, as follows: [5 7 2]. As
long as we agree that the vector components will be written in the order x, y, z,
we can use this notation instead of the unit-vector notation. In fact, MATLAB
uses this style for vector notation. MATLAB allows us to separate the components with commas for improved readability if we desire so that the equivalent
way of writing the preceding vector is [5, 7, 2]. This expression is a row vector,
which is a horizontal arrangement of the elements.
We can also express the vector as a column vector, which has a vertical
arrangement. A vector can have only one column, or only one row. Thus, a
vector is a one-dimensional array. In general, arrays can have more than one
column and more than one row.
Creating Vectors in MATLAB
The concept of a vector can be generalized to any number of components. In
MATLAB a vector is simply a list of scalars, whose order of appearance in the
list might be significant, as it is when specifying xyz coordinates. As another
example, suppose we measure the temperature of an object once every hour.
We can represent the measurements as a vector, and the 10th element in the
list is the temperature measured at the 10th hour.
To create a row vector in MATLAB, you simply type the elements inside a
pair of square brackets, separating the elements with a space or a comma. Brackets
are required for arrays unless you use the colon operator to create the array. In
this case you should not use brackets, but you can optionally use parentheses.
The choice between a space or comma is a matter of personal preference,
although the chance of an error is less if you use a comma. (You can also use a
comma followed by a space for maximum readability.)
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One- and Two-Dimensional Numeric Arrays
To create a column vector, you can separate the elements by semicolons;
alternatively, you can create a row vector and then use the transpose notation (’),
which converts a row vector into a column vector, or vice versa. For example:
>>g = [3;7;9]
g =
3
7
9
>>g = [3,7,9]’
g =
3
7
9
The third way to create a column vector is to type a left bracket ([) and the rst
element, press Enter, type the second element, press Enter, and so on until
you type the last element followed by a right bracket (]) and Enter. On the
screen this sequence looks like
>>g = [3
7
9]
g =
3
7
9
Note that MATLAB displays row vectors horizontally and column vectors
vertically.
You can create vectors by “appending” one vector to another. For example,
to create the row vector u whose rst three columns contain the values of r =
[2,4,20] and whose fourth, fth, and sixth columns contain the values of
w = [9,-6,3], you type u = [r,w]. The result is the vector u =
[2,4,20,9,-6,3].
The colon operator (:) easily generates a large vector of regularly spaced
elements. Typing
>>x = m:q:n
creates a vector x of values with a spacing q. The rst value is m. The last value
is n if m - n is an integer multiple of q. If not, the last value is less than n. For
example, typing x = 0:2:8 creates the vector x = [0,2,4,6,8], whereas
typing x = 0:2:7 creates the vector x = [0,2,4,6]. To create a row vector z consisting of the values from 5 to 8 in steps of 0.1, you type z = 5:0.1:8.
If the increment q is omitted, it is presumed to be 1. Thus y = -3:2 produces
the vector y = [-3,-2,-1,0,1,2].
55
TRANSPOSE
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The increment q can be negative. In this case m should be greater than n. For
example, u = 10:-2:4 produces the vector [10,8,6,4].
The linspace command also creates a linearly spaced row vector, but instead you specify the number of values rather than the increment. The syntax is
linspace(x1,x2,n), where x1 and x2 are the lower and upper limits and n
is the number of points. For example, linspace(5,8,31) is equivalent to
5:0.1:8. If n is omitted, the spacing is 1.
The logspace command creates an array of logarithmically spaced
elements. Its syntax is logspace(a,b,n), where n is the number of points
between 10a and 10b. For example, x = logspace(-1,1,4) produces the
vector x = [0.1000, 0.4642, 2.1544, 10.000]. If n is omitted, the
number of points defaults to 50.
Two-Dimensional Arrays
MATRIX
ARRAY SIZE
An array having rows and columns is a two-dimensional array that is sometimes
called a matrix. In mathematical text, if possible, vectors are usually denoted by
boldface lowercase letters and matrices by boldface uppercase letters. An example of a matrix having three rows and two columns is
2
M = -3
J
-7
5
4
K
1
We refer to the size of an array by the number of rows and the number of
columns. For example, an array with 3 rows and 2 columns is said to be a
3 2 array. The number of rows is always stated rst! We sometimes represent
a matrix A as [aij] to indicate its elements aij. The subscripts i and j, called
indices, indicate the row and column location of the element aij. The row number
must always come rst! For example, the element a32 is in row 3, column 2. Two
matrices A and B are equal if they have the same size and if all their corresponding elements are equal, that is, aij bij for every value of i and j.
Creating Matrices
The most direct way to create a matrix is to type the matrix row by row, separating the elements in a given row with spaces or commas and separating the rows
with semicolons. Brackets are required. For example, typing
>>A = [2,4,10;16,3,7];
creates the following matrix:
A = c
2
16
4
3
10
d
7
If the matrix has many elements, you can press Enter and continue typing on
the next line. MATLAB knows you are nished entering the matrix when you
type the closing bracket (]).
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2.1
One- and Two-Dimensional Numeric Arrays
You can append a row vector to another row vector to create either a third
row vector or a matrix (if both vectors have the same number of columns). Note
the difference between the results given by [a,b] and [a;b] in the following
session:
>>a = [1,3,5];
>>b = [7,9,11];
>>c = [a,b]
c =
1 3 5 7 9 11
>> D = [a;b]
D =
1 3 5
7 9 11
Matrices and the Transpose Operation
The transpose operation interchanges the rows and columns. In mathematics text
we denote this operation by the superscript T. For an m n matrix A with m
rows and n columns, AT (read “A transpose”) is an n m matrix.
A = c
-2
-3
6
d
5
AT = c
-2
6
-3
d
5
If AT A, the matrix A is symmetric. Note that the transpose operation converts
a row vector into a column vector, and vice versa.
If the array contains complex elements, the transpose operator (’) produces
the complex conjugate transpose; that is, the resulting elements are the complex
conjugates of the original array’s transposed elements. Alternatively, you can use
the dot transpose operator (.’) to transpose the array without producing complex conjugate elements, for example, A.’. If all the elements are real, the operators ‘ and.’ give the same result.
Array Addressing
Array indices are the row and column numbers of an element in an array and are used
to keep track of the array’s elements. For example, the notation v(5) refers to the
fth element in the vector v, and A(2,3) refers to the element in row 2, column 3
in the matrix A. The row number is always listed rst! This notation enables you to
correct entries in an array without retyping the entire array. For example, to change
the element in row 1, column 3 of a matrix D to 6, you can type D(1,3) = 6.
The colon operator selects individual elements, rows, columns, or “subarrays” of arrays. Here are some examples:
■
■
v(:) represents all the row or column elements of the vector v.
v(2:5) represents the second through fth elements; that is v(2), v(3),
v(4), v(5).
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■
■
■
■
■
■
A(:,3) denotes all the elements in the third column of the matrix A.
A(3,:) denotes all the elements in the third row of A.
A(:,2:5) denotes all the elements in the second through fth columns of A.
A(2:3,1:3) denotes all the elements in the second and third rows that
are also in the rst through third columns.
v = A(:) creates a vector v consisting of all the columns of A stacked
from rst to last.
A(end,:) denotes the last row in A, and A(:,end) denotes the last
column.
You can use array indices to extract a smaller array from another array. For example, if you create the array B
2
16
B = ≥
8
3
4
3
4
12
10
7
9
15
13
18
¥
25
17
(2.1–1)
by typing
>>B = [2,4,10,13;16,3,7,18;8,4,9,25;3,12,15,17];
and then type
>>C = B(2:3,1:3);
you can produce the following array:
C = c
EMPTY ARRAY
16
8
3
4
7
d
9
The empty array contains no elements and is expressed as []. Rows and
columns can be deleted by setting the selected row or column equal to the empty
array. This step causes the original matrix to collapse to a smaller one. For example,
A(3,:) = [] deletes the third row in A, while A(:,2:4) = [] deletes the
second through fourth columns in A. Finally, A([1 4],:) = [] deletes the rst
and fourth rows of A.
Suppose we type A = [6,9,4;1,5,7] to de ne the following matrix:
A = c
6
1
9
5
4
d
7
Typing A(1,5) = 3 changes the matrix to
A = c
6
1
9
5
4
7
0
0
3
d
0
Because A did not have ve columns, its size is automatically expanded to accept the new element in column 5. MATLAB adds zeros to ll out the remaining
elements.
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2.1
MATLAB does not accept negative or zero indices, but you can use negative
increments with the colon operator. For example, typing B = A(:,5:-1:1)
reverses the order of the columns in A and produces
B = c
3
0
0
0
4
7
9
5
6
d
1
Suppose that C = [-4,12,3,5,8]. Then typing B(2,:) = C replaces row 2
of B with C. Thus B becomes
B = c
3
-4
0
12
4
3
9
5
6
d
8
Suppose that D = [3,8,5;4,-6,9]. Then typing E = D([2,2,2],:)
repeats row 2 of D three times to obtain
4
E = 4
J
4
Using clear to Avoid Errors
-6
-6
-6
9
9
K
9
You can use the clear command to protect yourself from accidentally reusing an
array that has the wrong dimension. Even if you set new values for an array, some
previous values might still remain. For example, suppose you had previously created
the 2 2 array A = [2, 5; 6, 9], and then you create the 5 1 arrays x =
(1:5)’ and y = (2:6)’. Note that parentheses are needed here to use the transpose operator. Suppose you now rede ne A so that its columns will be x and y. If you
then type A(:,1) = x to create the rst column, MATLAB displays an error message telling you that the number of rows in A and x must be the same. MATLAB
thinks A should be a 2 2 matrix because A was previously de ned to have only two
rows and its values remain in memory. The clear command wipes A and all other
variables from memory and avoids this error. To clear A only, type clear A before
typing A(:,1) = x.
Some Useful Array Functions
MATLAB has many functions for working with arrays (see Table 2.1–1). Here is
a summary of some of the more commonly used functions.
The max(A) function returns the algebraically greatest element in A if A is
a vector having all real elements. It returns a row vector containing the greatest
elements in each column if A is a matrix containing all real elements. If any of
the elements are complex, max(A) returns the element that has the largest magnitude. The syntax [x,k] = max(A) is similar to max(A), but it stores the
maximum values in the row vector x and their indices in the row vector k.
If A and B have the same size, C = max(A,B) creates an array the same
size, having the maximum value from each corresponding location in A and B.
For example, the following A and B matrices give the C matrix shown.
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60
Table 2.1–1 Basic syntax of array functions*
Command
nd(x)
[u,v,w] =
Description
nd(A)
length(A)
linspace(a,b,n)
logspace(a,b,n)
max(A)
[x,k] = max(A)
min(A)
[x,k] = min(A)
norm(x)
size(A)
sort(A)
sum(A)
Computes an array containing the indices of the nonzero elements of the array x.
Computes the arrays u and v, containing the row and column indices of the nonzero
elements of the matrix A, and the array w, containing the values of the nonzero
elements. The array w may be omitted.
Computes either the number of elements of A if A is a vector or the largest value of
m or n if A is an m n matrix.
Creates a row vector of n regularly spaced values between a and b.
Creates a row vector of n logarithmically spaced values between a and b.
Returns the algebraically largest element in A if A is a vector. Returns a row vector
containing the largest elements in each column if A is a matrix. If any of the elements are complex, max(A) returns the elements that have the largest magnitudes.
Similar to max(A) but stores the maximum values in the row vector x and their
indices in the row vector k.
Same as max(A) but returns minimum values.
Same as [x,k] = max(A) but returns minimum values.
Computes a vector’s geometric length 2 x 12 + x22 + Á + x2n.
Returns a row vector [m n] containing the sizes of the m n array A.
Sorts each column of the array A in ascending order and returns an array the same
size as A.
Sums the elements in each column of the array A and returns a row vector containing the sums.
*Many of these functions have extended syntax. See the text and MATLAB help for more discussion.
A = c
1
3
6
7
4
d
2
B = c
3
1
4
5
7
d
8
C = c
3
3
6
7
7
d
8
The functions min(A) and [x,k] = min(A) are the same as max(A)
and [x,k] = max(A) except that they return minimum values.
The function size(A) returns a row vector [m n] containing the sizes of
the m n array A. The length(A) function computes either the number of elements of A if A is a vector or the largest value of m or n if A is an m n matrix.
For example, if
A =
6
- 10
J
3
2
-5
K
0
then max(A) returns the vector [6,2]; min(A) returns the vector [-10,-5];
size(A) returns [3,2]; and length(A) returns 3.
The sum(A) function sums the elements in each column of the array A and
returns a row vector containing the sums. The sort(A) function sorts each
column of the array A in ascending order and returns an array the same size as A.
If A has one or more complex elements, the max, min, and sort functions
act on the absolute values of the elements and return the element that has the largest
magnitude.
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2.1
For example, if
6
A =
- 10
J
3 + 4i
2
-5
K
0
6
A = 0
J
2
3
0
K
0
then max(A) returns the vector [-10,-5] and min(A) returns the vector
[3+4i,0]. (The magnitude of 3 4i is 5.)
The sort will be done in descending order if the form sort(A, ‘descend’)
is used. The min, max, and sort functions can be made to act on the rows instead of the columns by transposing the array.
The complete syntax of the sort function is sort(A, dim, mode),
where dim selects a dimension along which to sort and mode selects the direction of the sort, ‘ascend’ for ascending order and ‘descend’ for descending order. So, for example, sort(A,2, ‘descend’) would sort the
elements in each row of A in descending order.
The nd(x) command computes an array containing the indices of the
nonzero elements of the vector x. The syntax [u,v,w] = nd(A) computes
the arrays u and v, containing the row and column indices of the nonzero elements
of the matrix A, and the array w, containing the values of the nonzero elements.
The array w may be omitted.
For example, if
then the session
0
4
7
>>A = [6, 0, 3; 0, 4, 0; 2, 7, 0];
>>[u, v, w] = nd(A)
returns the vectors
1
3
u = ≥2¥
3
1
1
1
v = ≥2¥
2
3
6
2
w = ≥4¥
7
3
The vectors u and v give the (row, column) indices of the nonzero values,
which are listed in w. For example, the second entries in u and v give the indices
(3, 1), which speci es the element in row 3, column 1 of A, whose value is 2.
These functions are summarized in Table 2.1–1.
Magnitude, Length, and Absolute Value of a Vector
The terms magnitude, length, and absolute value are often loosely used in everyday
language, but you must keep their precise meaning in mind when using MATLAB.
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The MATLAB length command gives the number of elements in the vector. The
magnitude of a vector x having real elements x1, x2, . . . , xn is a
scalar, given by 2 x21 + x22 + Á + x2n, and is the same as the vector’s geometric length. The absolute value of a vector x is a vector whose elements are the
absolute values of the elements of x. For example, if x = [2,-4,5], its length
is 3; its magnitude is 222 + (- 4)2 + 52 = 6.7082; and its absolute value is
[2,4,5]. The length, magnitude, and absolute value of x are computed by
length(x), norm(x), and abs(x), respectively.
Test Your Understanding
T2.1–1 For the matrix B, nd the array that results from the operation [B;B’]. Use
MATLAB to determine what number is in row 5, column 3 of the result.
2
16
B = ≥
8
3
4
3
4
12
10
7
9
15
13
18
¥
25
17
T2.1–2 For the same matrix B, use MATLAB to (a) nd the lar gest and smallest
elements in B and their indices and (b) sort each column in B to
create a new matrix C.
The Variable Editor
The MATLAB Workspace Browser provides a graphical interface for managing
the workspace. You can use it to view, save, and clear workspace variables. It includes the Variable Editor, a graphical interface for working with variables,
including arrays. To open the Workspace Browser, type workspace at the
Command window prompt. The browser appears as shown in Figure 2.1–1.
Keep in mind that the Desktop menus are context-sensitive. Thus their
contents will change depending on which features of the browser and Variable
Editor you are currently using. The Workspace Browser shows the name of
each variable, its value, array size, and class. The icon for each variable illustrates its class.
Figure 2.1–1 The Workspace Browser.
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Figure 2.1–2 The Variable Editor.
From the Workspace Browser you can open the Variable Editor to view and
edit a visual representation of two-dimensional numeric arrays, with the rows and
columns numbered. To open the Variable Editor from the Workspace Browser,
double-click on the variable you want to open. The Variable Editor opens, displaying the values for the selected variable. The Variable Editor appears as shown
in Figure 2.1–2.
To open a variable, you can also right-click it and use the Context menu. Repeat the steps to open additional variables into the Variable Editor. In the Variable
Editor, access each variable via its tab at the bottom of the window, or use the
Window menu. You can also open the Variable Editor directly from the Command
window by typing open(‘var’), where var is the name of the variable to be
edited. Once an array is displayed in the Variable Editor, you can change a value in
the array by clicking on its location, typing in the new value, and pressing Enter.
Right-clicking on a variable brings up the Context menu, which can be used
to edit, save, or clear the selected variable, or to plot the rows of the variable versus its columns (this type of plot is discussed in Chapter 5).
You can also clear a variable from theWorkspace Browser by rst highlighting it in the Browser, then clicking on Delete in the Edit menu.
2.2 Multidimensional Numeric Arrays
MATLAB supports multidimensional arrays. For more information, type help
datatypes.
A three-dimensional array has the dimension m n q. A four-dimensional
array has the dimension m n q r, and so forth. The rst two dimensions are
the row and column, as with a matrix. The higher dimensions are called pages. You
can think of a three-dimensional array as layers of matrices. The rst layer is page 1;
the second layer is page 2, and so on. If A is a 3 3 2 array, you can access the
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element in row 3, column 2 of page 2 by typing A(3,2,2). To access all of
page 1, type A(:,:,1). To access all of page 2, type A(:,:,2). The ndims
command returns the number of dimensions. For example, for the array A just
described, ndims(A) returns the value 3.
You can create a multidimensional array by rst creating a two-dimensional
array and then extending it. For example, suppose you want to create a threedimensional array whose rst two pages are
4
5
J
3
6
8
9
1
0
K
2
6
0
J
4
2
3
7
9
1
K
5
To do so, rst create page 1 as a 3 3 matrix and then add page 2, as follows:
>>A = [4,6,1;5,8,0;3,9,2];
>>A(:,:,2) = [6,2,9;0,3,1;4,7,5];
Another way to produce such an array is with the cat command. Typing
cat(n,A,B,C,. ..) creates a new array by concatenating the arrays A, B,
C, and so on along the dimension n. Note that cat(1,A,B) is the same as
[A;B] and that cat(2,A,B) is the same as [A,B]. For example, suppose we
have the 2 2 arrays A and B:
4 6
8 2
B = c
d
d
9 5
7 3
Then C = cat(3,A,B) produces a three-dimensional array composed of two
layers; the rst layer is the matrix A, and the second layer is the matrix B. The
element C(m,n,p) is located in row m, column n, and layer p. Thus the element
C(2,1,1) is 9, and the element C(2,2,2) is 3.
Multidimensional arrays are useful for problems that involve several parameters. For example, if we have data on the temperature distribution in a rectangular
object, we could represent the temperatures as an array T with three dimensions.
A = c
2.3 Element-by-Element Operations
To increase the magnitude of a vector, multiply it by a scalar. For example, to
double the magnitude of the vector r = [3,5,2], multiply each component by
2 to obtain [6,10,4]. In MATLAB you type v = 2*r.
Multiplying a matrix A by a scalar w produces a matrix whose elements are
the elements of A multiplied by w. For example:
3c
2
5
9
6
d = c
-7
15
27
d
- 21
This multiplication is performed in MATLAB as follows:
>>A = [2,9;5,-7];
>>3*A
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Thus multiplication of an array by a scalar is easily de ned and easily carried out. However, multiplication of two arrays is not so straightforward. In fact,
MATLAB uses two de nitions of multiplication: (1) array multiplication and
(2) matrix multiplication. Division and exponentiation must also be carefully
de ned when you are dealing with operations between two arrays. MA TLAB has
two forms of arithmetic operations on arrays. In this section we introduce one
form, called array operations, which are also called element-by-element operations. In the next section we introduce matrix operations. Each form has its own
applications, which we illustrate by examples.
Array Addition and Subtraction
Array addition can be done by adding the corresponding components. To add the
arrays r = [3,5,2] and v = [2,-3,1] to create w in MATLAB, you type
w = r + v. The result is w = [5,2,3].
When two arrays have identical size, their sum or difference has the same
size and is obtained by adding or subtracting their corresponding elements. Thus
C A B implies that cij aij bij if the arrays are matrices. The array C has
the same size as A and B. For example,
c
6
10
-2
9
d + c
3
- 12
8
15
d = c
14
-2
6
d
17
(2.3–1)
Array subtraction is performed in a similar way.
The addition shown in Equation (2.3–1) is performed in MATLAB as follows:
>>A = [6,-2;10,3];
>>B = [9,8;-12,14]
>>A+B
ans =
15 6
-2 17
Array addition and subtraction are associative and commutative. For addition these properties mean that
(A + B) + C = A + (B + C)
(2.3–2)
A + B + C = B + C + A = A + C + B
(2.3–3)
Array addition and subtraction require that both arrays be the same size. The only
exception to this rule in MATLAB occurs when we add or subtract a scalar to or
from an array. In this case the scalar is added or subtracted from each element in
the array. Table 2.3–1 gives examples.
Element-by-Element Multiplication
MATLAB de nes element-by-element multiplication only for arrays that are the same
size. The de nition of the product x.*y, where x and y each have n elements, is
x.*y = [x(1)y(1), x(2)y(2) . . . , x(n)y(n)]
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Table 2.3–1 Element-by-element operations
Symbol
Operation
Form
Example
.*
./
.\
.^
Scalar-array addition
Scalar-array subtraction
Array addition
Array subtraction
Array multiplication
Array right division
Array left division
Array exponentiation
A b
A b
A B
A B
A.*B
A./B
A.\B
A.^B
[6,3]2[8,5]
[8,3]5[3,2]
[6,5][4,8][10,13]
[6,5][4,8][2,3]
[3,5].*[4,8][12,40]
[2,5]./[4,8][2/4,5/8]
[2,5].\[4,8][2\4,5\8]
[3,5].^2[3^2,5^2]
2.^[3,5][2^3,2^5]
[3,5].^[2,4][3^2,5^4]
if x and y are row vectors. For example, if
x = [2, 4, - 5]
y = [ -7, 3, - 8]
(2.3–4)
then z = x.*y gives
z = [2(- 7), 4(3), - 5(- 8)] = [- 14, 12, 40]
This type of multiplication is sometimes called array multiplication.
If u and v are column vectors, the result of u.*v is a column vector.
Note that x’ is a column vector with size 3 1 and thus does not have the
same size as y, whose size is 1 3. Thus for the vectors x and y the operations
x’.*y and y.*x’ are not de ned in MA TLAB and will generate an error message. With element-by-element multiplication, it is important to remember that
the dot (.) and the asterisk (*) form one symbol (.*). It might have been better
to have de ned a single symbol for this operation, but the developers of
MATLAB were limited by the selection of symbols on the keyboard.
The generalization of array multiplication to arrays with more than one row or
column is straightforward. Both arrays must be the same size. The array operations
are performed between the elements in corresponding locations in the arrays. For
example, the array multiplication operation A.*B results in a matrix C that has the
same size as A and B and has the elements cij aij bij. For example, if
-7
6
8
d
2
11(- 7) 5(8)
- 77
d = c
- 9(6) 4(2)
- 54
40
d
8
A = c
11
-9
5
d
4
B = c
then C = A.*B gives this result:
C = c
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Vectors and Displacement
Suppose two divers start at the surface and establish the following coordinate system: x is
to the west, y is to the north, and z is down. Diver 1 swims 55 ft west, 36 ft north, and then
dives 25 ft. Diver 2 dives 15 ft, then swims east 20 ft and then north 59 ft. (a) Find the distance between diver 1 and the starting point. (b) How far in each direction must diver 1
swim to reach diver 2? How far in a straight line must diver 1 swim to reach diver 2?
■ Solution
(a) Using the xyz coordinates selected, the position of diver 1 is r 55i 36j 25k, and
the position of diver 2 is r 20i 59j 15k. (Note that diver 2 swam east, which is
in the negative x direction.) The distance from the origin of a point xyz is given by
1 x2 + y2 + z2, that is, by the magnitude of the vector pointing from the origin to the
point xyz. This distance is computed in the following session.
>>r = [55,36,25];w = [-20,59,15];
>>dist1 = sqrt(sum(r.*r))
dist1 =
70.3278
The distance is approximately 70 ft. The distance could also have been computed from
norm(r).
(b) The location of diver 2 relative to diver 1 is given by the vector v pointing from
diver 1 to diver 2. We can nd this vector using vector subtraction: v w r. Continue
the above MATLAB session as follows:
>>v = w-r
v =
-75
23
-10
>>dist2 = sqrt(sum(v.*v))
dist2 =
79.0822
Thus to reach diver 2 by swimming along the coordinate directions, diver 1 must swim
75 ft east, 23 ft north, and 10 ft up. The straight-line distance between them is approximately 79 ft.
Vectorized Functions
The built-in MATLAB functions such as sqrt(x) and exp(x) automatically
operate on array arguments to produce an array result the same size as the array argument x. Thus these functions are said to be vectorized functions.
Thus, when multiplying or dividing these functions, or when raising them to a
power, you must use element-by-element operations if the arguments are arrays. For
example, to compute z (ey sin x) cos2 x, you must type z = exp(y).*
sin(x).*(cos(x)).^2. Obviously, you will get an error message if the size of
x is not the same as the size of y. The result z will have the same size as x and y.
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Aortic Pressure Model
EXAMPLE 2.3–2
The following equation is a speci c case of one model used to describe the blood pressure in the aorta during systole (the period following the closure of the heart’s aortic
valve). The variable t represents time in seconds, and the dimensionless variable y
represents the pressure difference across the aortic valve, normalized by a constant reference pressure.
y(t) = e-8t sin a 9.7t +
b
2
Plot this function for t 0.
■ Solution
Note that if t is a vector, the MATLAB functions exp(-8*t) and
sin(9.7*t+pi/2) will also be vectors the same size as t. Thus we must use
element-by-element multiplication to compute y(t).
We must decide on the proper spacing to use for the vector t and its upper limit.
The sine function sin(9.7t /2) oscillates with a frequency of 9.7 rad/sec, which is
9.7/(2) 1.5 Hz. Thus its period is 1/1.5 2/3 sec. The spacing of t should be a small
fraction of the period in order to generate enough points to plot the curve. Thus we select
a spacing of 0.003 to give approximately 200 points per period.
1.2
Normalized Pressure Difference y(t)
1
0.8
0.6
0.4
0.2
0
–0.2
0
0.05
0.1
0.15
0.2
0.25
t (sec)
0.3
0.35
Figure 2.3–1 Aortic pressure response for Example 2.3–2.
0.4
0.45
0.5
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The amplitude of the sine wave decays with time because the sine is multiplied by the
decaying exponential e8t. The exponential’s initial value is e0 1, and it will be 2 percent of its initial value at t 0.5 (because e8 (0.5) 0.02). Thus we select the upper limit
of t to be 0.5. The session is
>>t = 0:0.003:0.5;
>>y = exp(-8*t).*sin(9.7*t+pi/2);
>>plot(t,y),xlabel(‘t (sec)’), . . .
ylabel(‘Normalized Pressure Difference y(t)’)
The plot is shown in Figure 2.3–1. Note that we do not see much of an oscillation
despite the presence of a sine wave. This is so because the period of the sine wave is
greater than the time it takes for the exponential e8t to become essentially zero.
Element-by-Element Division
The de nition of element-by-element division, also called array division, is similar to the de nition of array multiplication except, of course, that the elements
of one array are divided by the elements of the other array. Both arrays must be
the same size. The symbol for array right division is . /. For example, if
x = [8, 12, 15]
y = [- 2, 6, 5]
then z = x./y gives
z = [8>( -2), 12>6, 15>5] = [ - 4, 2, 3]
Also, if
A = c
24
-9
20
d
4
-4
3
5
d
2
20>5
-6
d = c
4>2
-3
4
d
2
B = c
then C = A./B gives
C = c
24>( -4)
-9>3
The array left division operator (.\) is de ned to perform element-by-element
division using left division. Refer to Table 2.3–1 for examples. Note that A.\B is
not equivalent to A./B.
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Transportation Route Analysis
EXAMPLE 2.3–3
The following table gives data for the distance traveled along ve truck routes and the corresponding time required to traverse each route. Use the data to compute the average speed
required to drive each route. Find the route that has the highest average speed.
Distance (mi)
Time (hr)
1
2
3
4
5
560
10.3
440
8.2
490
9.1
530
10.1
370
7.5
■ Solution
For example, the average speed on the rst route is 560/10.3 54.4 mi/hr. First we dene the row vectors d and t from the distance and time data. Then, to nd the average
speed on each route using MATLAB, we use array division. The session is
>>d = [560, 440, 490, 530, 370]
>>t = [10.3, 8.2, 9.1, 10.1, 7.5]
>>speed = d./t
speed =
54.3689
53.6585
53.8462
52.4752
49.3333
The results are in miles per hour. Note that MATLAB displays more signi cant gures
than is justi ed by the three-signi cant- gure accuracy of the given data, so we should
round the results to three signi cant gures before using them.
To nd the highest average speed and the corresponding route, continue the session
as follows:
>>[highest_speed, route] = max(speed)
highest_speed =
54.3689
route =
1
The rst route has the highest speed.
If we did not need the speeds for every route, we could have solved this problem by
combining two lines as follows: [highest_speed, route] = max(d./t).
Element-by-Element Exponentiation
MATLAB enables us not only to raise arrays to powers but also to raise scalars
and arrays to array powers. To perform exponentiation on an element-byelement basis, we must use the .^ symbol. For example, if x = [3, 5, 8],
then typing x.^3 produces the array [33, 53, 83] [27, 125, 512]. If x =
0:2:6, then typing x.^2 returns the array [02, 22, 42, 62] [0, 4, 16, 36]. If
A = c
4
2
-5
d
3
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Element-by-Element Operations
71
then B = A.^3 gives this result:
64
43 ( - 5)3
B = c 3
3d = c
2
3
8
- 125
d
27
We can raise a scalar to an array power. For example, if p = [2, 4, 5],
then typing 3.^p produces the array [32, 34, 35] [9, 81, 243]. This example
illustrates a common situation in which it helps to remember that .^ is a single
symbol; the dot in 3.^p is not a decimal point associated with the number 3.
The following operations, with the value of p given here, are equivalent and
give the correct answer:
3.^p
3.0.^p
3..^p
(3).^p
3.^[2,4,5]
With array exponentiation, the power may be an array if the base is a scalar or if
the power’s dimensions are the same as the base dimensions. For example if, x =
[1,2,3] and y = [2,3,4], then y.^x gives the answer 2964. If A =
[1,2; 3,4], then 2.^A gives the array [2,4;8,16].
Test Your Understanding
T2.3–1 Given the matrices
21 27
–7 –3
d
B = c
d
- 18
8
9
4
nd (a) their array product, (b) their array right division ( A divided by
B), and (c) B raised to the third power element by element.
(Answers: (a) [-147, -81; -162, 32], (b) [-3, -9; -2,
2], and (c) [-343, -27; 729, 64].)
A = c
Current and Power Dissipation in Resistors
EXAMPLE 2.3–4
The current i passing through an electrical resistor having a voltage ␷ across it is given
by Ohm’s law, i = ␷/R, where R is the resistance. The power dissipated in the resistor is
given by ␷2/R. The following table gives data for the resistance and voltage for ve resistors. Use the data to compute (a) the current in each resistor and (b) the power dissipated
in each resistor.
1
R ()
␷ (V)
4
10
120
2
3
4
2 10
80
4
4
3.5 10
110
5
5
10
200
2 105
350
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■ Solution
(a) First we de ne two row vectors, one containing the resistance values and one containing
the voltage values. To nd the current i = ␷/R using MATLAB, we use array division.
The session is
>>R = [10000, 20000, 35000, 100000, 200000];
>>v = [120, 80, 110, 200, 350];
>>current = v./R
current =
0.0120 0.0040 0.0031 0.0020 0.0018
The results are in amperes and should be rounded to three signi cant gures because the
voltage data contains only three signi cant gures.
(b) To nd the power P = ␷2/R, use array exponentiation and array division. The
session continues as follows:
>>power = v.^2./R
power =
1.4400 0.3200
0.3457
0.4000
0.6125
These numbers are the power dissipation in each resistor in watts. Note that the statement
v.^2./R is equivalent to (v.^2)./R. Although the rules of precedence are unambiguous here, we can always put parentheses around quantities if we are unsure how
MATLAB will interpret our commands.
A Batch Distillation Process
EXAMPLE 2.3–5
Consider a system for heating a liquid benzene/toluene solution to distill a pure benzene
vapor. A particular batch distillation unit is charged initially with 100 mol of a 60 percent
mol benzene/40 percent mol toluene mixture. Let L (mol) be the amount of liquid remaining in the still, and let x (mol B/mol) be the benzene mole fraction in the remaining
liquid. Conservation of mass for benzene and toluene can be applied to derive the following relation [Felder, 1986].
L = 100 a
x 0.625 1 - x -1.625
a
b
b
0.6
0.4
Determine what mole fraction of benzene remains when L 70. Note that it is dif cult
to solve this equation directly for x. Use a plot of x versus L to solve the problem.
■ Solution
This equation involves both array multiplication and array exponentiation. Note that
MATLAB enables us to use decimal exponents to evaluate L. It is clear that L must be in
the range 0
L
100; however, we do not know the range of x, except that
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0.7
0.6
x (mol B/mol)
0.5
0.4
0.3
0.2
0.1
0
0
10
20
30
40
50
L (mol)
60
70
80
90
100
Figure 2.3–2 Plot for Example 2.3–5.
x 0. Therefore, we must make a few guesses for the range of x, using a session like the
following. We nd that L > 100 if x > 0.6, so we choose x = 0:0.001:0.6. We use
the ginput function to nd the value of x corresponding to L 70.
>>x = 0:0.001:0.6;
>>L = 100*(x/0.6).^(0.625).*((1-x)/0.4).^(-1.625);
>>plot(L,x),grid,xlabel(‘L(mol)’),ylabel(‘x (mol B/mol)’), ...
[L,x] = ginput(1)
The plot is shown in Figure 2.3–2. The answer is x 0. 52 if L 70. The
plot shows that the remaining liquid becomes leaner in benzene as the liquid amount becomes smaller. Just before the still is empty (L 0), the liquid is pure toluene.
2.4 Matrix Operations
Matrix addition and subtraction are identical to element-by-element addition and
subtraction. The corresponding matrix elements are summed or subtracted.
However, matrix multiplication and division are not the same as element-byelement multiplication and division.
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Multiplication of Vectors
Recall that vectors are simply matrices with one row or one column. Thus matrix
multiplication and division procedures apply to vectors as well, and we will introduce matrix multiplication by considering the vector case rst.
The vector dot product u ⭈ w of the vectors u and w is a scalar and can be
thought of as the perpendicular projection of u onto w. It can be computed from
|u||w| cos ␪, where ␪ is the angle between the two vectors and |u|, |w| are the magnitudes of the vectors. Thus if the vectors are parallel and in the same direction,
␪ ⫽ 0 and u ⭈ w ⫽ |u||w|. If the vectors are perpendicular, ␪ ⫽ 90° and thus u ⭈ w ⫽ 0.
Because the unit vectors i, j, and k have unit length,
i#i ⴝ j#j ⴝ k#k ⴝ 1
(2.4–1)
Because the unit vectors are perpendicular,
i#j ⴝ i#k ⴝ j#k ⴝ 0
(2.4–2)
Thus the vector dot product can be expressed in terms of unit vectors as
u # w = (u1i + u2 j + u3k) # (w1i + w2 j + w3k)
Carrying out the multiplication algebraically and using the properties given by
(2.4–1) and (2.4–2), we obtain
u # w = u1w1 + u2w2 + u3w3
The matrix product of a row vector u with a column vector w is de ned in
the same way as the vector dot product; the result is a scalar that is the sum of the
products of the corresponding vector elements; that is,
[u1
u2
w1
u3] w2 = u1w1 + u2w2 + u3w3
J K
w3
if each vector has three elements. Thus the result of multiplying a 1 ⫻ 3 vector by a 3 ⫻ 1 vector is a 1 ⫻ 1 array, that is, a scalar. This definition
applies to vectors having any number of elements, as long as both vectors
have the same number of elements.
Thus the result of multiplying a 1 ⫻ n vector by an n ⫻ 1 vector is a 1 ⫻ 1 array,
that is, a scalar.
Miles Traveled
EXAMPLE 2.4–1
Table 2.4–1 gives the speed of an aircraft on each leg of a certain trip and the time spent
on each leg. Compute the miles traveled on each leg and the total miles traveled.
■ Solution
We can de ne a row vector s containing the speeds and a row vector t containing the
times for each leg. Thus s = [200, 250, 400, 300] and t = [2, 5, 3, 4].
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Table 2.4–1 Aircraft speeds and times per leg
Leg
Speed (mi/hr)
Time (hr)
1
2
3
4
200
2
250
5
400
3
300
4
To nd the miles traveled on each leg, we multiply the speed by the time. To do so, we
use the MATLAB symbol .*, which speci es the multiplication s.*t to produce the
row vector whose elements are the products of the corresponding elements in s and t:
s.*t = [200(2), 250(5), 400(3), 300(4)] = [400, 1250, 1200, 1200]
This vector contains the miles traveled by the aircraft on each leg of the trip.
To nd the total miles traveled, we use the matrix product, denoted by s*t’. In this
de nition the product is the sum of the individual element products; that is,
s*t’ = [200(2) + 250(5) + 400(3) + 300(4)] = 4050
These two examples illustrate the difference between array multiplication s.*t and
matrix multiplication s*t’.
Vector-Matrix Multiplication
Not all matrix products are scalars. To generalize the preceding multiplication to
a column vector multiplied by a matrix, think of the matrix as being composed of
row vectors. The scalar result of each row-column multiplication forms an element in the result, which is a column vector. For example:
c
2
6
7 3
2(3) + 7(9)
69
d c d = c
d = c
d
-5 9
6(3) - 5(9)
- 27
(2.4–3)
Thus the result of multiplying a 2 2 matrix by a 2 1 vector is a 2 1 array,
that is, a column vector. Note that the de nition of multiplication requires that the
number of columns in the matrix be equal to the number of rows in the vector.
In general, the product Ax, where A has p columns, is de ned only if x has
p rows. If A has m rows and x is a column vector, the result of Ax is a
column vector with m rows.
Matrix-Matrix Multiplication
We can expand this de nition of multiplication to include the product of two
matrices AB. The number of columns in A must equal the number of rows in B. The
row-column multiplications form column vectors, and these column vectors form the
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matrix result. The product AB has the same number of rows as A and the same number of columns as B. For example,
6
10
J
4
(6)(9) + ( - 2)( - 5) (6)(8) + (-2)(12)
8
d = (10)(9) + (3)(- 5) (10)(8) + (3)(12)
12
J
K
(4)(9) + (7)(- 5)
(4)(8) + (7)(12)
-2
9
3 c
K 5
7
=
64
75
J1
24
116
K
116
(2.4–4)
Use the operator * to perform matrix multiplication in MATLAB. The following MATLAB session shows how to perform the matrix multiplication
shown in (2.4–4).
>>A = [6,-2;10,3;4,7];
>>B = [9,8;-5,12];
>>A*B
Element-by-element multiplication is de ned for the following product:
[3
1
7][4
6
5] = [12
6
35]
However, this product is not de ned for matrix multiplication, because the rst
matrix has three columns, but the second matrix does not have three rows. Thus
if we were to type [3, 1, 7]*[4, 6, 5] in MATLAB, we would receive
an error message.
The following product is de ned in matrix multiplication and gives the result shown:
x1
x1y1 x1y2
x2 [y1 y2 y3] = x2y1 x2y2
J x3 K
J x3y1 x3y2
The following product is also de ned:
[10 6] c
7
5
x1y3
x2y3
x3y3 K
4
d = [10(7) + 6(5) 10(4) + 6(2)] = [100 52]
2
Evaluating Multivariable Functions
To evaluate a function of two variables, say, z f (x, y), for the values x x1, x2,
. . . , xm and y y1, y2, . . . , yn, de ne the m n matrices:
y1 y2 Á yn
x1 x1 Á x1
x
x2 Á x2
y y2 Á yn
y = ≥ 1
x = ≥ 2
¥
¥
o
o
o
o
o
o
o
o
xm xm Á xm
y1 y2 Á yn
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77
When the function z f (x, y) is evaluated in MATLAB using array operations, the
resulting m n matrix z has the elements zij f (xi , yj). We can extend this technique
to functions of more than two variables by using multidimensional arrays.
Height versus Velocity
EXAMPLE 2.4–2
The maximum height h achieved by an object thrown with a speed y at an angle
horizontal, neglecting drag, is
to the
␷2 sin2
2g
Create a table showing the maximum height for the following values of y and :
h =
␷ = 10, 12, 14, 16, 18, 20 m/s
= 50°, 60°, 70°, 80°
The rows in the table should correspond to the speed values, and the columns should correspond to the angles.
■ Solution
The program is shown below.
g = 9.8; v = 10:2:20;
theta = 50:10:80;
h = (v’.ˆ2)*(sind(theta).ˆ2)/(2*g);
table = [0, theta; v’, h]
The arrays v and theta contain the given velocities and angles. The array v is 1 6
and the array theta is 1 4. Thus the term v’.ˆ2 is a 6 1 array, and the term
sind(theta).ˆ2 is a 1 4 array. The product of these two arrays, h, is a matrix product and is a (6 1)(1 4) (6 4) matrix.
The array [0, theta] is 1 5 and the array [v’, h] is 6 5, so the matrix
table is 7 5. The following table shows the matrix table rounded to one decimal place.
From this table we can see that the maximum height is 8.8 m if y 14 m/s and 70°.
0
50
60
70
80
10
12
14
16
18
20
3.0
4.3
5.9
7.7
9.7
12.0
3.8
5.5
7.5
9.8
12.4
15.3
4.5
6.5
8.8
11.5
14.6
18.0
4.9
7.1
9.7
12.7
16.0
19.8
Test Your Understanding
T2.4–1 Use MATLAB to compute the dot product of the following vectors:
u = 6i - 8j + 3k
w = 5i + 3j - 4k
Check your answer by hand. (Answer: 6.)
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T2.4–2 Use MATLAB to show that
7
-3
J
5
4
1
2 c
K 7
9
35
8
d = 11
6
J
68
80
- 12
K
94
Manufacturing Cost Analysis
EXAMPLE 2.4–3
Table 2.4–2 shows the hourly cost of four types of manufacturing processes. It also shows
the number of hours required of each process to produce three different products. Use
matrices and MATLAB to solve the following. (a) Determine the cost of each process to
produce 1 unit of product 1. (b) Determine the cost to make 1 unit of each product. (c) Suppose we produce 10 units of product 1, 5 units of product 2, and 7 units of product 3.
Compute the total cost.
■ Solution
(a) The basic principle we can use here is that cost equals the hourly cost times the
number of hours required. For example, the cost of using the lathe for product 1 is
($10/hr)(6 hr) $60, and so forth for the other three processes. If we de ne the row vector of hourly costs to be hourly_costs and de ne the row vector of hours required for
product 1 to be hours_1, then we can compute the costs of each process for product 1
using element-by-element multiplication. In MATLAB the session is
>>hourly_cost = [10, 12, 14, 9];
>>hours_1 = [6, 2, 3, 4];
>>process_cost_1 = hourly_cost.*hours_1
process_cost_1 =
60
24
42
36
These are the costs of each of the four processes to produce 1 unit of product 1.
(b) To compute the total cost of 1 unit of product 1, we can use the vectors
hourly_costs and hours_1 but apply matrix multiplication instead of element-byelement multiplication, because matrix multiplication sums the individual products. The
matrix multiplication gives
Table 2.4–2 Cost and time data for manufacturing processes
Hours required to produce one unit
Process
Hourly cost ($)
Product 1
Product 2
Product 3
Lathe
Grinding
Milling
Welding
10
12
14
9
6
2
3
4
5
3
2
0
4
1
5
3
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6
2
[10 12 14 9] ≥ ¥ = 10(6) + 12(2) + 14(3) + 9(4) = 162
3
4
We can perform similar multiplication for products 2 and 3, using the data in the table.
For product 2:
5
3
[10 12 14 9] ≥ ¥ = 10(5) + 12(2) + 14(3) + 9(0) = 114
2
0
For product 3:
4
1
[10 12 14 9] ≥ ¥ = 10(4) + 12(1) + 14(5) + 9(3) = 149
5
3
These three operations could have been accomplished in one operation by de ning a
matrix whose columns are formed by the data in the last three columns of the table:
6 5 4
60 + 24 + 42 + 36
2 3 1
¥ = 50 + 36 + 28 + 0 = [162 114 149]
[10 12 14 9] ≥
3 2 5
J 40 + 12 + 70 + 27 K
4 0 3
In MATLAB the session continues as follows. Remember that we must use the transpose
operation to convert the row vectors into column vectors.
>>hours_2 = [5, 3, 2, 0];
>>hours_3 = [4, 1, 5, 3];
>>unit_cost = hourly_cost*[hours_1’, hours_2’, hours_3’]
unit_cost =
162
114
149
Thus the costs to produce 1 unit each of products 1, 2, and 3 are $162, $114, and $149,
respectively.
(c) To nd the total cost to produce 10, 5, and 7 units, respectively , we can use matrix multiplication:
162
[10 5 7] 114 = 1620 + 570 + 1043 = 3233
J
K
149
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In MATLAB the session continues as follows. Note the use of the transpose operator on
the vector unit_cost.
>>units = [10, 5, 7];
>>total_cost = units*unit_cost’
total_cost =
3233
The total cost is $3233.
The General Matrix Multiplication Case
We can state the general result for matrix multiplication as follows: Suppose A
has dimension m p and B has dimension p q. If C is the product AB, then C
has dimension m q and its elements are given by
p
cij = a aik bkj
(2.4–5)
k=1
for all i 1, 2, . . . , m and j 1, 2, . . . , q. For the product to be de ned, the matrices A and B must be conformable; that is, the number of rows in B must equal
the number of columns in A. The product has the same number of rows as A and
the same number of columns as B.
Matrix multiplication does not have the commutative property; that is, in
general, AB Z BA. Reversing the order of matrix multiplication is a common and
easily made mistake.
The associative and distributive properties hold for matrix multiplication.
The associative property states that
A(B + C) ⴝ AB + AC
(2.4–6)
The distributive property states that
(AB)C ⴝ A(BC)
(2.4–7)
Applications to Cost Analysis
Project cost data stored in tables must often be analyzed in several ways. The elements in MATLAB matrices are similar to the cells in a spreadsheet, and MATLAB
can perform many spreadsheet-type calculations for analyzing such tables.
EXAMPLE 2.4–4
Product Cost Analysis
Table 2.4–3 shows the costs associated with a certain product, and Table 2.4–4 shows the
production volume for the four quarters of the business year. Use MATLAB to nd the
quarterly costs for materials, labor, and transportation; the total material, labor, and transportation costs for the year; and the total quarterly costs.
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Table 2.4–3 Product costs
Unit costs ($ 103)
Product
Materials
Labor
Transportation
1
2
3
4
6
2
4
9
2
5
3
7
1
4
2
3
Table 2.4–4 Quarterly production volume
Product
Quarter 1
Quarter 2
Quarter 3
Quarter 4
1
2
3
4
10
8
12
6
12
7
10
4
13
6
13
11
15
4
9
5
■ Solution
The costs are the product of the unit cost and the production volume. Thus we de ne two
matrices: U contains the unit costs in Table 2.4–3 in thousands of dollars, and P contains
the quarterly production data in Table 2.4–4.
>>U = [6, 2, 1;2, 5, 4;4, 3, 2;9, 7, 3];
>>P = [10, 12, 13, 15;8, 7, 6, 4;12, 10, 13, 9;6, 4, 11, 5];
Note that if we multiply the rst column in U by the rst column in P, we obtain the
total materials cost for the rst quarter . Similarly, multiplying the rst column in U by the
second column in P gives the total materials cost for the second quarter. Also, multiplying the second column in U by the rst column in P gives the total labor cost for the rst
quarter, and so on. Extending this pattern, we can see that we must multiply the transpose
of U by P. This multiplication gives the cost matrix C.
>>C = U’*P
The result is
C =
178
138
J
84
162
117
72
241
172
96
179
112
K
64
Each column in C represents one quarter. The total rst-quarter cost is the sum of the elements in the rst column, the second-quarter cost is the sum of the second column, and so
on. Thus because the sum command sums the columns of a matrix, the quarterly costs are
obtained by typing
>>Quarterly_Costs = sum(C)
The resulting vector, containing the quarterly costs in thousands of dollars, is [400 351 509
355]. Thus the total costs in each quarter are $400,000; $351,000; $509,000; and $355,000.
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The elements in the rst row of C are the material costs for each quarter; the elements in the second row are the labor costs, and those in the third row are the transportation costs. Thus to nd the total material costs, we must sum across the rst row of C.
Similarly, the total labor and total transportation costs are the sums across the second and
third rows of C. Because the sum command sums columns, we must use the transpose of
C. Thus we type the following:
>>Category_Costs ⫽ sum(C’)
The resulting vector, containing the category costs in thousands of dollars, is [760 539
316]. Thus the total material costs for the year are $760 000; the labor costs are $539 000;
and the transportation costs are $316 000.
We displayed the matrix C only to interpret its structure. If we need not display C,
the entire analysis would consist of only four command lines.
>>U = [6, 2, 1;2, 5, 4;4, 3, 2;9, 7, 3];
>>P = [10, 12, 13, 15;8, 7, 6, 4;12, 10, 13, 9;6, 4, 11, 5];
>>Quarterly_Costs = sum(U’*P)
Quarterly_Costs =
400
351
509
355
>>Category_Costs = sum((U’*P)’)
Category_Costs =
760
539
316
This example illustrates the compactness of MATLAB commands.
Special Matrices
NULL MATRIX
IDENTITY MATRIX
Two exceptions to the noncommutative property are the null matrix, denoted by
0, and the identity, or unity, matrix, denoted by I. The null matrix contains all
zeros and is not the same as the empty matrix [ ], which has no elements. The
identity matrix is a square matrix whose diagonal elements are all equal to 1,
with the remaining elements equal to 0. For example, the 2 ⫻ 2 identity matrix is
I = c
1
0
0
d
1
These matrices have the following properties:
0A ⴝ A0 ⴝ 0
IA ⴝ AI ⴝ A
MATLAB has speci c commands to create several special matrices. Type
help specmat to see the list of special matrix commands; also check Table 2.4–5.
The identity matrix I can be created with the eye(n) command, where n is
the desired dimension of the matrix. To create the 2 ⫻ 2 identity matrix, you type
eye(2). Typing eye(size(A)) creates an identity matrix having the same
dimension as the matrix A.
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Table 2.4–5 Special matrices
Command
Description
eye(n)
eye(size(A))
ones(n)
ones(m,n)
ones(size(A))
zeros(n)
zeros(m,n)
zeros(size(A))
Creates an n n identity matrix.
Creates an identity matrix the same size as the matrix A.
Creates an n n matrix of 1s.
Creates an m n array of 1s.
Creates an array of 1s the same size as the array A.
Creates an n n matrix of 0s.
Creates an m n array of 0s.
Creates an array of 0s the same size as the array A.
Sometimes we want to initialize a matrix to have all zero elements. The
zeros command creates a matrix of all zeros. Typing zeros(n) creates an
n n matrix of zeros, whereas typing zeros(m,n) creates an m n matrix of
zeros, as will typing A(m,n) = 0. Typing zeros(size(A)) creates a
matrix of all zeros having the same dimension as the matrix A. This type of
matrix can be useful for applications in which we do not know the required
dimension ahead of time. The syntax of the ones command is the same, except
that it creates arrays lled with 1s.
For example, to create and plot the function
f(x) =
the script le is
10 0 … x … 2
0 2 6 x 6 5
L -3 5 … x … 7
x1 = 0:0.01:2;
f1 = 10*ones(size(x1));
x2 = 2.01:0.01:4.99;
f2 = zeros(size(x2));
x3 = 5:0.01:7;
f3 = -3*ones(size(x3));
f = [f1, f2, f3];
x = [x1, x2, x3];
plot(x,f),xlabel(‘x’),ylabel(‘y’)
(Consider what the plot would look like if the command plot(x,f) were replaced with the command plot(x1,f1,x2,f2,x3,f3).)
Matrix Division and Linear Algebraic Equations
Matrix division uses both the right and left division operators, / and \, for various
applications, a principal one being the solution of sets of linear algebraic equations. Chapter 8 covers a related topic, the matrix inverse.
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You can use the left division operator (\) in MATLAB to solve sets of linear
algebraic equations. For example, consider the set
6x + 12y + 4z = 70
7x - 2y + 3z = 05
2x + 8y - 9z = 64
To solve such sets in MATLAB you must create two arrays; we will call them
A and B. The array A has as many rows as there are equations and as many
columns as there are variables. The rows of A must contain the coef cients of
x, y, and z in that order. In this example, the rst row of A must be 6, 12, 4; the
second row must be 7, 2, 3; and the third row must be 2, 8, 9. The array B
contains the constants on the right-hand side of the equation; it has one column
and as many rows as there are equations. In this example, the rst row of B is
70, the second is 5, and the third is 64. The solution is obtained by typing A\B.
The session is
>>A = [6,12,4;7,-2,3;2,8,-9];
>>B = [70;5;64];
>>Solution = A\B
Solution =
3
5
-2
LEFT DIVISION
METHOD
The solution is x 3, y 5, and z 2.
The left division method works ne when the equation set has a unique
solution. To learn how to deal with problems having a nonunique solution (or
perhaps no solution at all!), see Chapter 8.
Test Your Understanding
T2.4–3 Use MATLAB to solve the following set of equations.
6x - 4y + 8z = 112
- 5x - 3y + 7z =
75
14x + 9y - 5z = - 67
(Answer: x 2, y 5, z 10.)
Matrix Exponentiation
Raising a matrix to a power is equivalent to repeatedly multiplying the matrix by
itself, for example, A2 AA. This process requires the matrix to have the same
number of rows as columns; that is, it must be a square matrix. MATLAB uses
the symbol ^ for matrix exponentiation. To nd A2, type A^2.
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We can raise a scalar n to a matrix power A, if A is square, by typing n^A,
but the applications for such a procedure are in advanced courses. However,
raising a matrix to a matrix power—that is, AB—is not de ned, even if A and B
are square.
Special Products
Many applications in physics and engineering use the cross product and dot
product; for example, calculations to compute moments and force components use these special products. If A and B are vectors with three elements,
the cross product command cross(A,B) computes the three-element vector
that is the cross product A B. If A and B are 3 n matrices, cross(A,B)
returns a 3 n array whose columns are the cross products of the corresponding columns in the 3 n arrays A and B. For example, the moment M with
respect to a reference point O due to the force F is given by M r F, where
r is the position vector from the point O to the point where the force F is applied. To find the moment in MATLAB, you type M = cross(r,F).
The dot product command dot(A,B) computes a row vector of length n
whose elements are the dot products of the corresponding columns of the m n
arrays A and B. To compute the component of the force F along the direction
given by the vector r, you type dot(F,r).
2.5 Polynomial Operations Using Arrays
MATLAB has some convenient tools for working with polynomials. Type help
polyfun for more information on this category of commands. We will use the
following notation to describe a polynomial:
f(x) = a1x n + a2x n - 1 + a3x n - 2 + Á + an - 1x 2 + an x + a n + 1
We can describe a polynomial in MATLAB with a row vector whose elements
are the polynomial’s coef cients, starting with the coef cient of the highest
power of x. This vector is [a1, a2, a3, . . . , an1, an, an1]. For example, the vector
[4,-8,7,-5] represents the polynomial 4x3 8x2 7x 5.
Polynomial roots can be found with the roots(a) function, where a is the
array containing the polynomial coef cients. For example, to obtain the roots of
x3 12x2 45x 50 0, you type y = roots([1,12,45,50]). The answer (y) is a column array containing the values 2, 5, 5.
The poly(r) function computes the coef cients of the polynomial whose
roots are speci ed by the array r. The result is a row array that contains the polynomial’s coef cients. For example, to nd the polynomial whose roots are 1 and
3 5i, the session is
>>p = poly([1,3+5i, 3-5i])
p =
1
-7 40 -34
Thus the polynomial is x3 7x2 40x 34.
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Polynomial Addition and Subtraction
To add two polynomials, add the arrays that describe their coef cients. If the
polynomials are of different degrees, add zeros to the coef cient array of the
lower-degree polynomial. For example, consider
f (x) 9x3 5x2 3x 7
whose coef cient array is f [9,5,3,7] and
g(x) 6x2 x 2
whose coef cient array is g = [6,-1,2]. The degree of g(x) is 1 less that of
f (x). Therefore, to add f (x) and g(x), we append one zero to g to “fool”
MATLAB into thinking g(x) is a third-degree polynomial. That is, we
type g = [0 g] to obtain [0,6,-1,2] for g. This vector represents g(x) 0x3 6x2 x 2. To add the polynomials, type h = f+g. The result is
h = [9,1,2,9], which corresponds to h(x) 9x3 x2 2x 9. Subtraction is done in a similar way.
Polynomial Multiplication and Division
To multiply a polynomial by a scalar, simply multiply the coef cient array by
that scalar. For example, 5h(x) is represented by [45,5,10,45].
Multiplication and division of polynomials are easily done with MATLAB.
Use the conv function (it stands for “convolve”) to multiply polynomials
and use the deconv function (deconv stands for “deconvolve”) to perform
synthetic division. Table 2.5–1 summarizes these functions, as well as the poly,
polyval, and roots functions.
Table 2.5–1 Polynomial functions
Command
Description
conv(a,b)
Computes the product of the two polynomials described by the coef cient arrays a and b.
The two polynomials need not be of the same degree. The result is the coef cient array
of the product polynomial.
Computes the result of dividing a numerator polynomial, whose coef cient array is num,
by a denominator polynomial represented by the coef cient array den. The quotient
polynomial is given by the coef cient array q, and the remainder polynomial is given by
the coef cient array r.
Computes the coef cients of the polynomial whose roots are speci ed by the vector r. The
result is a row vector that contains the polynomial’s coef cients arranged in descending
order of power.
Evaluates a polynomial at speci ed values of its independent variable x, which can be a
matrix or a vector. The polynomial’s coef cients of descending powers are stored in the
array a. The result is the same size as x.
Computes the roots of a polynomial speci ed by the coef cient array a. The result is a
column vector that contains the polynomial’s roots.
[q,r] deconv (num,den)
poly(r)
polyval(a,x)
roots(a)
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The product of the polynomials f(x) and g(x) is
f (x)g(x) (9x 3 5x2 3x 7)(6x2 x 2)
54x5 39x4 41x3 29x2 x 14
Dividing f(x) by g(x) using synthetic division gives a quotient of
f(x)
9x3 - 5x2 + 3x + 7
= 1.5x - 0.5833
=
g(x)
6x2 - x + 2
with a remainder of 0.5833x 8.1667. Here is the MATLAB session to perform these operations.
>>f = [9,-5,3,7];
>>g = [6,-1,2];
>>product = conv(f,g)
product =
54
-39
41
29
-1
14
>>[quotient, remainder] = deconv(f,g)
quotient =
1.5
-0.5833
remainder =
0
0
-0.5833
8.1667
The conv and deconv functions do not require that the polynomials be of the
same degree, so we did not have to fool MATLAB as we did when adding
the polynomials.
Plotting Polynomials
The polyval(a,x) function evaluates a polynomial at speci ed values of its
independent variable x, which can be a matrix or a vector. The polynomial’s
coef cient array is a. The result is the same size as x. For example, to evaluate
the polynomial f(x) 9x3 5x2 3x 7 at the points x 0, 2, 4, . . . , 10, type
>>f = polyval([9,-5,3,7],[0:2:10]);
The resulting vector f contains six values that correspond to f(0), f(2), f(4), . . . , f(10).
The polyval function is very useful for plotting polynomials. To do this,
you should de ne an array that contains many values of the independent variable
x in order to obtain a smooth plot. For example, to plot the polynomial f(x) 9x3 5x2 3x 7 for 2 x 5, you type
>>x = -2:0.01:5;
>>polyval([9,-5,3,7], x);
>>plot (x,f),xlabel( x ),ylabel( f(x) ),grid
Polynomial derivatives and integrals are covered in Chapter 9.
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Test Your Understanding
T2.5–1 Use MATLAB to obtain the roots of
x3 13x2 52x 6 0
Use the poly function to con rm your answer .
T2.5–2 Use MATLAB to con rm that
(20x3 7x2 5x 10)(4x2 12x 3)
80x5 212x4 124x3 121x2 105x 30
T2.5–3 Use MATLAB to con rm that
12x3 + 5x2 - 2x + 3
3x2 - 7x + 4
with a remainder of 59x 41.
= 4x + 11
T2.5–4 Use MATLAB to con rm that
6x3 + 4x2 - 5
12x3 - 7x2 + 3x + 9
= 0.7108
when x 2.
T2.5–5 Plot the polynomial
y x3 13x2 52x 6
over the range 7
EXAMPLE 2.5–1
x
1.
Earthquake-Resistant Building Design
Buildings designed to withstand earthquakes must have natural frequencies of vibration
that are not close to the oscillation frequency of the ground motion. A building’s natural
frequencies are determined primarily by the masses of its oors and by the lateral stif fness of its supporting columns (which act as horizontal springs). We can nd these frequencies by solving for the roots of a polynomial called the structure’s characteristic
polynomial (characteristic polynomials are discussed further in Chapter 9). Figure 2.5–1
shows the exaggerated motion of the oors of a three-story building. For such a building,
if each oor has a mass m and the columns have stiffness k, the polynomial is
( - f2)[(2 - f2)2 - 2] + 2f2 - 23
where = k>4m2 (models such as these are discussed in greater detail in [Palm,
2010]). The building’s natural frequencies in cycles per second are the positive roots of
this equation. Find the building’s natural frequencies in cycles per second for the case
where m = 1000 kg and k = 5 * 106 N/m.
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Floor
Floor
Columns
Floor
Ground Motion
Figure 2.5–1 Simple vibration model of a building subjected to
ground motion.
■ Solution
The characteristic polynomial consists of sums and products of lower-degree polynomials. We can use this fact to have MATLAB do the algebra for us. The characteristic polynomial has the form
p1 A p22 - 2 B + p3 = 0
where
p1 = - f2
p2 = 2 - f2
p3 = 2f2 - 23
The MATLAB script le is
k = 5e+6;m = 1000;
alpha = k/(4*m*pi^2);
p1 = [-1,0,alpha];
p2 = [-1,0,2*alpha];
p3 = [alpha^2,0,-2*alpha^3];
p4 = conv(p2,p2)-(0,0,0,0,alpha^2];
p5 = conv(p1,p4);
p6 = p5+[0,0,0,0,p3];
r = roots(p6)
The resulting positive roots and thus the frequencies, rounded to the nearest integer, are
20, 14, and 5 Hz.
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2.6 Cell Arrays
The cell array is an array in which each element is a bin, or cell, which can contain an array. You can store different classes of arrays in a cell array, and you can
group data sets that are related but have different dimensions. You access cell
arrays using the same indexing operations used with ordinary arrays.
This is the only section in the text that uses cell arrays. Coverage of this section is therefore optional. Some more advanced MATLAB applications, such as
those found in some of the toolboxes, do use cell arrays.
Creating Cell Arrays
CELL INDEXING
CONTENT
INDEXING
You can create a cell array by using assignment statements or by using the
cell function. You can assign data to the cells by using either cell indexing
or content indexing. To use cell indexing, enclose in parentheses the cell subscripts on the left side of the assignment statement and use the standard array
notation. Enclose the cell contents on the right side of the assignment statement
in braces {}.
An Environment Database
EXAMPLE 2.6–1
Suppose you want to create a 2 2 cell array A, whose cells contain the location, the
date, the air temperature (measured at 8 A.M., 12 noon, and 5 P.M.), and the water temperatures measured at the same time in three different points in a pond. The cell array
looks like the following.
Walden Pond
[60
■ Solution
72
65]
June 13, 1997
55
54
J 52
57
56
55
56
55
53 K
You can create this array by typing the following either in interactive mode or in a script
le and running it.
A(1,1) = {‘Walden Pond’};
A(1,2) = {‘June 13, 1997’};
A(2,1) = {[60,72,65]};
A(2,2) = {[55,57,56;54,56,55;52,55,53]};
If you do not yet have contents for a particular cell, you can type a pair of empty
braces { } to denote an empty cell, just as a pair of empty brackets [] denotes an empty
numeric array. This notation creates the cell but does not store any contents in it.
To use content indexing, enclose in braces the cell subscripts on the left side, using the
standard array notation. Then specify the cell contents on the right side of the assignment
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operator. For example:
A{1,1} = ‘Walden Pond’;
A{1,2} = ‘June 13, 1997’;
A{2,1} = [60,72,65];
A{2,2} = [55,57,56;54,56,55;52,55,53];
Type A at the command line. You will see
A ‘Walden Pond’
[1x3 double]
‘June 13, 1997’
[3x3 double]
You can use the celldisp function to display the full contents. For example, typing
celldisp(A) displays
A{1,1} =
Walden Pond
A{2,1} =
60 72 65
.
.
.
etc.
The cellplot function produces a graphical display of the cell array’s
contents in the form of a grid. Type cellplot(A) to see this display for the
cell array A. Use commas or spaces with braces to indicate columns of cells and
use semicolons to indicate rows of cells (just as with numeric arrays). For example, typing
B = {[2,4], [6,-9;3,5]; [7;2], 10};
creates the following 2 2 cell array:
[2
4] c
[7
2]
6
3
-9
d
5
10
You can preallocate empty cell arrays of a speci ed size by using the cell function. For example, type C = cell(3,5) to create the 3 5 cell array C and
ll it with empty matrices. Once the array has been de ned in this way , you
can use assignment statements to enter the contents of the cells. For example,
type C(2,4) = {[6,-3,7]} to put the 1 3 array in cell (2,4) and type
C(1,5) = {1:10} to put the numbers from 1 to 10 in cell (1,5).
Type C(3,4) = {‘30 mph’} to put the string in cell (3,4).
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Accessing Cell Arrays
You can access the contents of a cell array by using either cell indexing or content
indexing. To use cell indexing to place the contents of cell (3,4) of the array C in
the new variable Speed, type Speed = C(3,4). To place the contents of
the cells in rows 1 to 3, columns 2 to 5 in the new cell array D, type D =
C(1:3,2:5). The new cell array D will have three rows, four columns, and
12 arrays. To use content indexing to access some of or all the contents in a
single cell, enclose the cell index expression in braces to indicate that you are
assigning the contents, not the cells themselves, to a new variable. For example,
typing Speed = C{3,4} assigns the contents ‘30 mph’ in cell (3,4) to the
variable Speed. You cannot use content indexing to retrieve the contents of
more than one cell at a time. For example, the statements G = C{1,:} and
C{1,:} = var, where var is some variable, are both invalid.
You can access subsets of a cell’s contents. For example, to obtain the second element in the 1 3-row vector in the (2,4) cell of array C and assign it to the
variable r, you type r = C{2,4}(1,2). The result is r = -3.
2.7 Structure Arrays
FIELD
Structure arrays are composed of structures. This class of arrays enables you to
store dissimilar arrays together. The elements in structures are accessed using
named elds. This feature distinguishes them from cell arrays, which are
accessed using the standard array indexing operations.
Structure arrays are used in this text only in this section. Some MATLAB
toolboxes do use structure arrays.
A speci c example is the best way to introduce the terminology of structures.
Suppose you want to create a database of students in a course, and you want to include each student’s name, Social Security number, email address, and test scores.
Figure 2.7–1 shows a diagram of this data structure. Each type of data (name,
Social Security number, and so on) is a eld, and its name is the eld name. Thus
our database has four elds. The rst three elds each contain a text string, while
Structure array “student”
Student(1)
Student(2)
Name: John Smith
Name: Mary Jones
SSN: 392-77-1786
SSN: 431-56-9832
Email: smithj@myschool.edu
Email: jonesm@myschool.edu
Tests: 67, 75, 84
Tests: 84, 78, 93
Figure 2.7–1 Arrangement of data in the structure array student.
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93
the last eld (the test scores) contains a vector having numerical elements. A structure consists of all this information for a single student. A structure array is an
array of such structures for different students. The array shown in Figure 2.7–1 has
two structures arranged in one row and two columns.
Creating Structures
You can create a structure array by using assignment statements or by using the
struct function. The following example uses assignment statements to build a
structure. Structure arrays use the dot notation (.) to specify and to access the elds.
You can type the commands either in the interactive mode or in a script le.
A Student Database
Create a structure array to contain the following types of student data:
■
■
■
■
Student name.
Social Security number.
Email address.
Test scores.
Enter the data shown in Figure 2.7–1 into the database.
■ Solution
You can create the structure array by typing the following either in the interactive mode
or in a script le. Start with the data for the rst student.
student.name = ‘John Smith’;
student.SSN = ‘392-77-1786’;
student.email = ‘smithj@myschool.edu’;
student.tests = [67,75,84];
If you then type
>>student
at the command line, you will see the following response:
name: ‘John Smith’
SSN: = ‘392-77-1786’
email: = ‘smithj@myschool.edu’
tests: = [67 75 84]
To determine the size of the array, type size(student). The result is ans 1 1,
which indicates that it is a 1 1 structure array.
To add a second student to the database, use a subscript 2 enclosed in parentheses
after the structure array’s name and enter the new information. For example, type
student(2).name = ‘Mary Jones’;
student(2).SSN = ‘431-56-9832’;
student(2).email = ‘jonesm@myschool.edu’;
student(2).tests = [84,78,93];
EXAMPLE 2.7–1
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This process “expands” the array. Before we entered the data for the second student, the
dimension of the structure array was 1 1 (it was a single structure). Now it is a 1 2 array
consisting of two structures, arranged in one row and two columns. You can con rm this
information by typing size(student), which returns ans 1 2. If you now type
length(student), you will get the result ans = 2, which indicates that the array has
two elements (two structures). When a structure array has more than one structure, MATLAB
does not display the individual eld contents when you type the structure array’s name. For
example, if you now type student, MATLAB displays
>>student =
1x2 struct array with
name
SSN
email
tests
elds:
You can also obtain information about the elds by using the
Table 2.7–1). For example:
eldnames function (see
>> eldnames(student)
ans =
‘name’
‘SSN’
‘email’
‘tests’
As you ll in more student information, MA TLAB assigns the same number of elds and
the same eld names to each element. If you do not enter some information—for
example, suppose you do not know someone’s email address—MATLAB assigns an
empty matrix to that eld for that student.
Table 2.7–1 Structure functions
Function
names =
Description
eldnames(S)
is eld(S,’ eld’)
isstruct(S)
S = rm eld(S,’ eld’)
S = struct(‘f1’,’v1’,’f2’,
’v2’, ...)
Returns the eld names associated
with the structure array S as
names, a cell array of strings.
Returns 1 if ‘ eld’ is the
name of a eld in the structure
array S and 0 otherwise.
Returns 1 if the array S is a
structure array and 0 otherwise.
Removes the eld ‘ eld’
from the structure array S.
Creates a structure array with the
elds ‘f1’, ‘f2’, . . . having
the values ‘v1’, ‘v2’, . . . .
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The elds can he dif ferent sizes. For example, each name eld can contain a dif ferent number of characters, and the arrays containing the test scores can be different sizes,
as would be the case if a certain student did not take the second test.
In addition to the assignment statement, you can build structures using the
struct function, which lets you “preallocate” a structure array. To build a
structure array named sa_1, the syntax is
sa_1 = struct(‘ eld1’,’values1’,’ eld2’,values2’, . . .)
where the arguments are the eld names and their values. The values arrays
values1, values2, . . . must all be arrays of the same size, scalar cells, or
single values. The elements of the values arrays are inserted into the corresponding elements of the structure array. The resulting structure array has the same size
as the values arrays, or is 1 1 if none of the values arrays is a cell. For example,
to preallocate a 1 1 structure array for the student database, you type
student = struct(‘name’,’John Smith’, ‘SSN’, . . .
‘392-77-1786’,’email’,’smithj@myschool.edu’, . . .
‘tests’,[67,75,84])
Accessing Structure Arrays
To access the contents of a particular eld, type a period after the structure array
name, followed by the eld name. For example, typing student(2).name
displays the value ‘Mary Jones’. Of course, we can assign the result to a
variable in the usual way. For example, typing name2 = student(2).name
assigns the value ‘Mary Jones’ to the variable name2. To access elements
within a eld, for example, John Smith’ s second test score, type student(1).
tests(2). This entry returns the value 75. In general, if a eld contains an array ,
you use the array’s subscripts to access its elements. In this example the statement
student(1).tests(2) is equivalent to student(1,1).tests(2)
because student has one row.
To store all the information for a particular structure—say, all the information about Mary Jones—in another structure array named M, you type M =
student(2). You can also assign or change values of field elements. For
example, typing student(2).tests(2) = 81 changes Mary Jones’s
second test score from 78 to 81.
Modifying Structures
Suppose you want to add phone numbers to the database. You can do this by typing the rst student’ s phone number as follows:
student(1).phone = ‘555-1653’
All the other structures in the array will now have a phone eld, but these elds
will contain the empty array until you give them values.
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To delete a eld from every structure in the array , use the rm eld function.
Its basic syntax is
new_struc = rm eld(array,’ eld’);
where array is the structure array to be modi ed, ‘ eld’ is the eld to be
removed, and new_struc is the name of the new structure array so created by
the removal of the eld. For example, to remove the Social Security eld and call
the new structure array new_student, type
new_student = rm eld(student,’SSN’);
Using Operators and Functions with Structures
You can apply the MATLAB operators to structures in the usual way. For
example, to nd the maximum test score of the second student, you type
max(student(2).tests). The answer is 93.
The is eld function determines whether a structure array contains a particular eld. Its syntax is is eld(S, ‘ eld’). It returns a value of 1 (which
means “true”) if ‘ eld’ is the name of a eld in the structure array S. For example, typing is eld(student, ‘name’) returns the result ans = 1.
The isstruct function determines whether an array is a structure array.
Its syntax is isstruct(S). It returns a value of 1 if S is a structure array and
0 otherwise. For example, typing isstruct(student) returns the result
ans = 1, which is equivalent to “true.”
Test Your Understanding
T2.7–1 Create the structure array student shown in Figure 2.7–1 and add the
following information about a third student: name: Alfred E. Newman;
SSN: 555-12-3456; e-mail: newmana@myschool.edu; tests: 55, 45, 58.
T2.7–2 Edit your structure array to change Newman’s second test score from
45 to 53.
T2.7–3 Edit your structure array to remove the SSN eld.
2.8 Summary
You should now be able to perform basic operations and use arrays in MATLAB.
For example, you should be able to
■
■
■
Create, address, and edit arrays.
Perform array operations including addition, subtraction, multiplication,
division, and exponentiation.
Perform matrix operations including addition, subtraction, multiplication,
division, and exponentiation.
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Problems
Table 2.8–1 Guide to commands introduced in Chapter 2
Special
characters
’
.’
;
:
Use
Transposes a matrix, creating complex
conjugate elements.
Transposes a matrix without creating complex
conjugate elements.
Suppresses screen printing; also denotes a new
row in an array.
Represents an entire row or column of an array.
Tables
Array functions
Element-by-element operations
Special matrices
Polynomial functions
Structure functions
■
■
Table 2.1–1
Table 2.3–1
Table 2.4–5
Table 2.5–1
Table 2.7–1
Perform polynomial algebra.
Create databases using cell and structure arrays.
Table 2.8–1 is a reference guide to all the MATLAB commands introduced in
this chapter.
Key Terms with Page References
Absolute value, 61
Array addressing, 57
Array operations, 65
Array size, 56
Cell array, 90
Cell indexing, 90
Column vector, 54
Content indexing, 90
Empty array, 58
Field, 92
Identity matrix, 82
Left division method, 84
Length, 61
Magnitude, 61
Matrix, 56
Matrix operations, 73
Null matrix, 82
Row vector, 54
Structure arrays, 92
Transpose, 55
Problems
You can nd the answers to problems marked with an asterisk at the end of the text.
Section 2.1
1.
a. Use two methods to create the vector x having 100 regularly spaced
values starting at 5 and ending at 28.
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b. Use two methods to create the vector x having a regular spacing of 0.2
starting at 2 and ending at 14.
c. Use two methods to create the vector x having 50 regularly spaced values starting at 2 and ending at 5.
2.
a. Create the vector x having 50 logarithmically spaced values starting at
10 and ending at 1000.
b. Create the vector x having 20 logarithmically spaced values starting at
10 and ending at 1000.
3.*
Use MATLAB to create a vector x having six values between 0 and 10
(including the endpoints 0 and 10). Create an array A whose rst row
contains the values 3x and whose second row contains the values
5x 20.
4.
Repeat Problem 3 but make the rst column of A contain the values 3x
and the second column contain the values 5x 20.
5.
Type this matrix in MATLAB and use MATLAB to carry out the following
instructions.
3
-5
A = ≥
6
15
7
9
13
5
-4
10
8
4
12
2
¥
11
1
a. Create a vector v consisting of the elements in the second column of A.
b. Create a vector w consisting of the elements in the second row of A.
6.
Type this matrix in MATLAB and use MATLAB to carry out the following
instructions.
3
7
-5
9
A = ≥
6 13
15
5
- 4 12
10
2
¥
8 11
4
1
a. Create a 4 3 array B consisting of all elements in the second
through fourth columns of A.
b. Create a 3 4 array C consisting of all elements in the second
through fourth rows of A.
c. Create a 2 3 array D consisting of all elements in the rst two rows
and the last three columns of A.
7.*
Compute the length and absolute value of the following vectors:
a. x [2, 4, 7]
b. y [2, 4, 7]
c. z [5 3i, 3 4i, 2 7i]
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Problems
8.
Given the matrix
3
-5
A = ≥
6
15
7
9
13
5
- 4 12
10
2
¥
8 11
4
1
a. Find the maximum and minimum values in each column.
b. Find the maximum and minimum values in each row.
9.
Given the matrix
3
-5
A = ≥
6
15
7
9
13
5
-4
10
8
4
12
2
¥
11
1
a. Sort each column and store the result in an array B.
b. Sort each row and store the result in an array C.
c. Add each column and store the result in an array D.
d. Add each row and store the result in an array E.
10.
Consider the following arrays.
1
2
A = ≥
7
3
4
4
9
2
100
¥
7
42
B = ln(A)
Write MATLAB expressions to do the following.
a. Select just the second row of B.
b. Evaluate the sum of the second row of B.
c. Multiply the second column of B and the rst column of A element by
element.
d. Evaluate the maximum value in the vector resulting from element-byelement multiplication of the second column of B with the rst column
of A.
e. Use element-by-element division to divide the rst row of A by the
rst three elements of the third column of B. Evaluate the sum of the
elements of the resulting vector.
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Section 2.2
11.* a. Create a three-dimensional array D whose three “layers” are these
matrices:
3
A = 6
J
7
-2
8
9
1
-5
K
10
6
B =
7
J -8
9
5
2
-4
3
1K
-7
C = 10
J 3
-5
6
-9
-5
d
-2
C = c
-3
7
-9
d
8
2
1
8K
b. Use MATLAB to nd the lar gest element in each layer of D and the
largest element in D.
Section 2.3
12.* Given the matrices
- 7 11
4
d
B = c
4
9
12
Use MATLAB to
a. Find A B C.
b. Find A B C.
c. Verify the associative law
A = c
(A + B) + C ⴝ A + (B + C)
d. Verify the commutative law
A + B + C ⴝ B + C + A ⴝ A + C + B
13.* Given the matrices
A = c
56
24
32
d
- 16
B = c
14
6
-4
d
-2
Use MATLAB to
a. Find the result of A times B using the array product.
b. Find the result of A divided by B using array right division.
c. Find B raised to the third power element by element.
14.* The mechanical work W done in using a force F to push a block through a
distance D is W FD. The following table gives data on the amount of
force used to push a block through the given distance over ve segments
of a certain path. The force varies because of the differing friction properties of the surface.
Path segment
Force (N)
Distance (m)
1
2
3
4
5
400
3
550
0.5
700
0.75
500
1.5
600
5
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Problems
Use MATLAB to nd ( a) the work done on each segment of the path and
(b) the total work done over the entire path.
15.
Plane A is heading southwest at 300 mi/hr, while plane B is heading west
at 150 mi/hr. What are the velocity and the speed of plane A relative to
plane B?
16.
The following table shows the hourly wages, hours worked, and output
(number of widgets produced) in one week for ve widget makers.
Worker
Hourly wage ($)
Hours worked
Output (widgets)
1
2
3
4
5
5
40
1000
5.50
43
1100
6.50
37
1000
6
50
1200
6.25
45
1100
Use MATLAB to answer these questions:
a. How much did each worker earn in the week?
b. What is the total salary amount paid out?
c. How many widgets were made?
d. What is the average cost to produce one widget?
e. How many hours does it take to produce one widget on average?
f. Assuming that the output of each worker has the same quality, which
worker is the most ef cient? Which is the least ef cient?
17.
Two divers start at the surface and establish the following coordinate system: x is to the west, y is to the north, and z is down. Diver 1 swims 60 ft
east, then 25 ft south, and then dives 30 ft. At the same time, diver 2 dives
20 ft, swims east 30 ft and then south 55 ft.
a. Compute the distance between diver 1 and the starting point.
b. How far in each direction must diver 1 swim to reach diver 2?
c. How far in a straight line must diver 1 swim to reach diver 2?
18.
The potential energy stored in a spring is kx2/2, where k is the spring constant and x is the compression in the spring. The force required to compress the spring is kx. The following table gives the data for ve springs:
Spring
Force (N)
Spring constant k (N/m)
1
2
3
4
5
11
1000
7
600
8
900
10
1300
9
700
Use MATLAB to nd ( a) the compression x in each spring and (b) the potential energy stored in each spring.
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19.
A company must purchase ve kinds of material. The following table
gives the price the company pays per ton for each material, along with the
number of tons purchased in the months of May, June, and July:
Quantity purchased (tons)
Material
Price ($/ton)
May
June
July
300
550
400
250
500
5
3
6
3
2
4
2
5
5
4
6
4
3
4
3
1
2
3
4
5
Use MATLAB to answer these questions:
a. Create a 53 matrix containing the amounts spent on each item for
each month.
b. What is the total spent in May? in June? in July?
c. What is the total spent on each material in the three-month period?
d. What is the total spent on all materials in the three-month period?
20.
A fenced enclosure consists of a rectangle of length L and width 2R, and a
semicircle of radius R, as shown in Figure P20. The enclosure is to be
built to have an area A of 1600 ft2. The cost of the fence is $40/ft for the
curved portion and $30/ft for the straight sides. Use the min function to
determine with a resolution of 0.01 ft the values of R and L required to
minimize the total cost of the fence. Also compute the minimum cost.
L
2R
R
Figure P20
21.
A water tank consists of a cylindrical part of radius r and height h, and a
hemispherical top. The tank is to be constructed to hold 500 m3 of uid
when lled. The surface area of the cylindrical part is 2rh, and its volume is r2 h. The surface area of the hemispherical top is given by 2r2,
and its volume is given by 2r3/3. The cost to construct the cylindrical
part of the tank is $300/m2 of surface area; the hemispherical part costs
$400/m2. Plot the cost versus r for 2 r 10 m, and determine the
radius that results in the least cost. Compute the corresponding height h.
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Problems
22.
Write a MATLAB assignment statement for each of the following functions, assuming that w, x, y, and z are row vectors of equal length and that
c and d are scalars.
f =
1
12c/x
A =
23.
E =
e-c>(2x)
(ln y) 2dz
x + w/(y + z)
x + w/(y - z)
S =
x(2.15 + 0.35y)1.8
z(1 - x)y
a. After a dose, the concentration of medication in the blood declines due
to metabolic processes. The half-life of a medication is the time required after an initial dosage for the concentration to be reduced by
one-half. A common model for this process is
C(t) = C(0)e-kt
where C(0) is the initial concentration, t is time (in hours), and k is
called the elimination rate constant, which varies among individuals.
For a particular bronchodilator, k has been estimated to be in the range
0.047 k 0.107 per hour. Find an expression for the half-life in
terms of k, and obtain a plot of the half-life versus k for the indicated
range.
b. If the concentration is initially zero and a constant delivery rate is
started and maintained, the concentration as a function of time is
described by
C(t) =
24.
a
(1 - e-kt)
k
where a is a constant that depends on the delivery rate. Plot the concentration after 1 hr, C (1), versus k for the case where a 1 and k is
in the range 0.047 k 0.107 per hour.
A cable of length Lc supports a beam of length Lb , so that it is horizontal when the weight W is attached at the beam end. The principles of
statics can be used to show that the tension force T in the cable is
given by
T =
LbLcW
D 2L2b - D2
where D is the distance of the cable attachment point to the beam pivot.
See Figure P24.
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Lc
W
D
Lb
Figure P24
a. For the case where W 400 N, Lb 3 m, and Lc 5 m, use element-byelement operations and the min function to compute the value of D that
minimizes the tension T. Compute the minimum tension value.
b. Check the sensitivity of the solution by plotting T versus D. How
much can D vary from its optimal value before the tension T increases
10 percent above its minimum value?
Section 2.4
25.* Use MATLAB to nd the products AB and BA for the following
matrices:
A = c
26.
11
-9
5
d
-4
B = c
-7
6
-8
d
2
Given the matrices
4
A = 6
J
7
-2
8
9
1
-5
10 K
6
7
B =
J
-8
Use MATLAB to
a. Verify the associative property
9
5
2
-4
3
K
1
C =
-4
10
J
3
-5
6
-9
2
1
K
8
A(B + C) ⴝ AB + AC
b. Verify the distributive property
(AB)C ⴝ A(BC)
27.
The following tables show the costs associated with a certain product and
the production volume for the four quarters of the business year. Use MATLAB to nd ( a) the quarterly costs for materials, labor, and transportation;
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Problems
(b) the total material, labor, and transportation costs for the year; and (c) the
total quarterly costs.
Unit product costs ($ 103)
Product
Materials
Labor
Transportation
1
2
3
4
5
7
3
9
2
6
3
1
4
5
2
2
3
5
4
1
Quarterly production volume
Product
Quarter 1
Quarter 2
Quarter 3
Quarter 4
1
2
3
4
5
16
12
8
14
13
14
15
9
13
16
10
11
7
15
12
12
13
11
17
18
28.* Aluminum alloys are made by adding other elements to aluminum to improve its properties, such as hardness or tensile strength. The following
table shows the composition of ve commonly used alloys, which are
known by their alloy numbers (2024, 6061, and so on) [Kutz, 1999]. Obtain
a matrix algorithm to compute the amounts of raw materials needed to produce a given amount of each alloy. Use MATLAB to determine how much
raw material of each type is needed to produce 1000 tons of each alloy.
Composition of aluminum alloys
29.
Alloy
%Cu
%Mg
%Mn
%Si
%Zn
2024
6061
7005
7075
356.0
4.4
0
0
1.6
0
1.5
1
1.4
2.5
0.3
0.6
0
0
0
0
0
0.6
0
0
7
0
0
4.5
5.6
0
Redo Example 2.4–4 as a script le to allow the user to examine the effects of labor costs. Allow the user to input the four labor costs in the following table. When you run the le, it should display the quarterly costs
and the category costs. Run the le for the case where the unit labor costs
are $3000, $7000, $4000, and $8000, respectively.
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Product costs
Unit costs ($ 103)
Product
Materials
Labor
Transportation
1
2
3
4
6
2
4
9
2
5
3
7
1
4
2
3
Quarterly production volume
30.
Product
Quarter 1
Quarter 2
Quarter 3
Quarter 4
1
2
3
4
10
8
12
6
12
7
10
4
13
6
13
11
15
4
9
5
Vectors with three elements can represent position, velocity, and acceleration. A mass of 5 kg, which is 3 m away from the x axis, starts at
x 2 m and moves with a speed of 10 m/s parallel to the y axis. Its
velocity is thus described by v [0, 10, 0], and its position is described
by r [2, 10t 3, 0]. Its angular momentum vector L is found from
L m(r v), where m is the mass. Use MATLAB to
a. Compute a matrix P whose 11 rows are the values of the position vector r evaluated at the times t 0, 0.5, 1, 1.5, . . . , 5 s.
b. What is the location of the mass when t 5 s?
c. Compute the angular momentum vector L. What is its direction?
31.* The scalar triple product computes the magnitude M of the moment of a
force vector F about a speci ed line. It is M (r F) . n, where r is the
position vector from the line to the point of application of the force and n
is a unit vector in the direction of the line.
Use MATLAB to compute the magnitude M for the case where
F [12, 5, 4] N, r [ 3, 5, 2] m, and n [6, 5, 7].
32.
Verify the identity
A : (B : C) ⴝ B (A # C) ⴚ C(A # B)
for the vectors A ⫽ 7i 3j 7k, B ⫽ 6i 2j 3k, and
C ⫽ 2i 8j 8k.
33.
The area of a parallelogram can be computed from |A ⫻ B|, where A and
B de ne two sides of the parallelogram (see Figure P33). Compute the
area of a parallelogram de ned by A ⫽ 5i and B ⫽ i 3j.
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Problems
y
B
A
x
Figure P33
34.
The volume of a parallelepiped can be computed from |A (B ⫻C)|,
where A, B, and C de ne three sides of the parallelepiped (see Figure P34). Compute the volume of a parallelepiped de ned by A ⫽ 5i,
B ⫽ 2i 4j, and C ⫽ 3i 2k.
y
B
C
A
x
z
Figure P34
Section 2.5
35.
Use MATLAB to plot the polynomials y 3x4 6x3 8x2 4x 90
and z 3x3 5x2 8x 70 over the interval 3 x 3. Properly
label the plot and each curve. The variables y and z represent current in
milliamperes; the variable x represents voltage in volts.
36.
Use MATLAB to plot the polynomial y 3x4 5x3 28x2 5x 200
on the interval 1 x 1. Put a grid on the plot and use the ginput
function to determine the coordinates of the peak of the curve.
37.
Use MATLAB to nd the following product:
(10x3 - 9x2 - 6x + 12)(5x3 - 4x2 - 12x + 8)
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38.* Use MATLAB to nd the quotient and remainder of
14x3 - 6x2 + 3x + 9
5x2 + 7x - 4
39.* Use MATLAB to evaluate
24x3 - 9x2 - 7
10x3 + 5x2 - 3x - 7
at x ⫽ 5.
40.
The ideal gas law provides one way to estimate the pressures and volumes of a gas in a container. The law is
RT
N
V
More accurate estimates can be made with the van der Waals equation
P =
P =
41.
RT
-
a
N2
V
N - b
V
where the term b is a correction for the volume of the molecules and the
term a/V̂ 2 is a correction for molecular attractions. The values of a and b
depend on the type of gas. The gas constant is R, the absolute temperature
is T, and the gas speci c volume is V̂ . If 1 mol of an ideal gas were conned to a volume of 22.41 L at 0⬚C (273.2 K), it would exert a pressure of
1 atm. In these units, R ⫽ 0.08206.
For chlorine (Cl2), a ⫽ 6.49 and b ⫽ 0.0562. Compare the speci c
volume estimates V̂ given by the ideal gas law and the van der Waals
equation for 1 mol of Cl2 at 300 K and a pressure of 0.95 atm.
Aircraft A is ying east at 320 mi/hr , while aircraft B is ying south at
160 mi/hr. At 1:00 P.M. the aircraft are located as shown in Figure P41.
B
160 mi/h
410 mi
A
320 mi/h
800 mi
Figure P41
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Problems
a. Obtain the expression for the distance D between the aircraft as a function of time. Plot D versus time until D reaches its minimum value.
b. Use the roots function to compute the time when the aircraft are rst
within 30 mi of each other.
42.
The function
y =
3x2 - 12x + 20
x2 - 7x + 10
approaches ⬁ as x → 2 and as x → 5. Plot this function over the range 0 ⱕ
x ⱕ 7. Choose an appropriate range for the y axis.
43.
The following formulas are commonly used by engineers to predict the
lift and drag of an airfoil:
L =
1
␳CLSV2
2
D =
1
␳C SV2
2 D
where L and D are the lift and drag forces, V is the airspeed, S is the wing
span, ␳ is the air density, and CL and CD are the lift and drag coef cients.
Both CL and CD depend on ␣, the angle of attack, the angle between the
relative air velocity and the airfoil’s chord line.
Wind tunnel experiments for a particular airfoil have resulted in the
following formulas.
CL = 4.47 * 10-5␣3 + 1.15 * 10-3␣2 + 6.66 * 10-2␣ + 1.02 * 10-1
CD = 5.75 * 10-6␣3 + 5.09 * 10-4␣2 + 1.8 * 10-4␣ + 1.25 * 10-2
44.
where ␣ is in degrees.
Plot the lift and drag of this airfoil versus V for 0 ⱕ V ⱕ 150 mi/hr
(you must convert V to ft /sec; there is 5280 ft/mi). Use the values ␳ ⫽
0.002378 slug/ft3 (air density at sea level), ␣ ⫽ 10°, and S ⫽ 36 ft. The resulting values of L and D will be in pounds.
The lift-to-drag ratio is an indication of the effectiveness of an airfoil. Referring to Problem 43, the equations for lift and drag are
L =
1
␳C SV 2
2 L
D =
1
␳C SV 2
2 D
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where, for a particular airfoil, the lift and drag coef cients versus angle of
attack are given by
CL = 4.47 * 10-53 + 1.15 * 10-32 + 6.66 * 10-2 + 1.02 * 10-1
CD = 5.75 * 10-63 + 5.09 * 10-42 + 1.81 * 10-4 + 1.25 * 10-2
Using the rst two equations, we see that the lift-to-drag ratio is given
simply by the ratio CL /CD .
1
2
CL
L
2 CLSV
= 1
=
D
CD
CDSV 2
2
Plot L/D versus for 2°
maximizes L/D.
22°. Determine the angle of attack that
Section 2.6
45.
a. Use both cell indexing and content indexing to create the following
2 2 cell array.
Motor 28C
c
3
7
9
d
2
Test ID 6
[6 5 1]
b. What are the contents of the (1,1) element in the (2,1) cell in this array?
46.
The capacitance of two parallel conductors of length L and radius r, separated by a distance d in air, is given by
C =
L
ln[(d - r)/r]
where is the permittitivity of air ( 8.854 1012 F/m). Create a cell
array of capacitance values versus d, L, and r for d 0.003, 0.004, 0.005,
and 0.01 m; L 1, 2, 3 m; and r 0.001, 0.002, 0.003 m. Use MATLAB
to determine the capacitance value for d 0.005, L 2, and r 0.001.
Section 2.7
47.
a. Create a structure array that contains the conversion factors for converting units of mass, force, and distance between the metric SI system
and the British Engineering System.
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Problems
b. Use your array to compute the following:
■
■
■
■
■
■
48.
The number of meters in 48 ft.
The number of feet in 130 m.
The number of pounds equivalent to 36 N.
The number of newtons equivalent to 10 lb.
The number of kilograms in 12 slugs.
The number of slugs in 30 kg.
Create a structure array that contains the following information elds concerning the road bridges in a town: bridge location, maximum load (tons),
year built, year due for maintenance. Then enter the following data into
the array:
Location
Max. load
Year built
Due for maintenance
Smith St.
Hope Ave.
Clark St.
North Rd.
80
90
85
100
1928
1950
1933
1960
2011
2013
2012
2012
49.
Edit the structure array created in Problem 48 to change the maintenance
data for the Clark St. bridge from 2012 to 2018.
50.
Add the following bridge to the structure array created in Problem 48.
Location
Max. load
Year built
Due for maintenance
Shore Rd.
85
1997
2014
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Courtesy Jet Propulsion Lab/NASA.
Engineering in the
21st Century. . .
Robot-Assisted Microsurgery
Y
ou need not be a medical doctor to participate in the exciting developments now taking place in the health eld. Many advances in medicine
and surgery are really engineering achievements, and many engineers
are contributing their talents in this area. Recent achievements include
■
■
■
Laparoscopic surgery in which a ber -optic scope guides a small surgical
device. This technology eliminates the need for large incisions and the resulting long recuperation.
Computerized axial tomography (CAT) scans and magnetic resonance
imaging (MRI), which provide noninvasive tools for diagnosing medical
problems.
Medical instrumentation, such as a ngertip sensor for continuously measuring oxygen in the blood and automatic blood pressure sensors.
As we move into the 21st century, an exciting challenge will be the development
of robot-assisted surgery in which a robot, supervised by a human, performs operations requiring precise, steady motions. Robots have already assisted in hip
surgery on animals, but much greater development is needed. Another developing technology is telesurgery in which a surgeon uses a television interface to remotely guide a surgical robot. This technology would allow delivery of medical
services to remote areas.
Robot-assisted microsurgery, which uses a robot capable of very small,
precise motions, shows great promise. In some applications the robotic device is
used to lter out tremors normally present in the human hand. Designing such
devices requires geometric analysis, control system design, and image processing. The MATLAB Image Processing toolbox and the several MATLAB toolboxes dealing with control system design are useful for such applications. ■
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C H A P T E R
3
Functions and Files
OUTLINE
3.1 Elementary Mathematical Functions
3.2 User-De ned Functions
3.3 Additional Function Topics
3.4 Working with Data Files
3.5 Summary
Problems
MATLAB has many built-in functions, including trigonometric, logarithmic, and
hyperbolic functions, as well as functions for processing arrays. These functions
are summarized in Section 3.1. In addition, you can de ne your own functions
with a function le, and you can use them just as conveniently as the built-in functions. We explain this technique in Section 3.2. Section 3.3 covers additional topics in function programming, including function handles, anonymous functions,
subfunctions, and nested functions. Another type of le that is useful in MA TLAB
is the data le. Importing and exporting such les is covered in Section 3.4.
Sections 3.1 and 3.2 contain essential topics and must be covered. The
material in Section 3.3 is useful for creating large programs. The material in Section 3.4 is useful for readers who must work with large data sets.
3.1 Elementary Mathematical Functions
You can use the lookfor command to nd functions that are relevant to your
application. For example, type lookfor imaginary to get a list of the functions that deal with imaginary numbers. You will see listed
imag Complex imaginary part
i
Imaginary unit
j
Imaginary unit
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Note that imaginary is not a MATLAB function, but the word is found in
the Help descriptions of the MATLAB function imag and the special symbols
i and j. Their names and brief descriptions are displayed when you type
lookfor imaginary. If you know the correct spelling of a MATLAB function, for example, disp, you can type help disp to obtain a description
of the function.
Some of the functions, such as sqrt and sin, are built in. These are stored
as image les and are not M- les. They are part of the MATLAB core so they are
very ef cient, but the computational details are not readily accessible. Some
functions are implemented in M- les. You can see the code and even modify it,
although this is not recommended.
Exponential and Logarithmic Functions
Table 3.1–1 summarizes some of the common elementary functions. An example
is the square root function sqrt. To compute 19, you type sqrt(9) at the
command line. When you press Enter, you see the result ans = 3. You can use
functions with variables. For example, consider the session
>>x = -9; y = sqrt(x)
y =
0 + 3.0000i
Note that the sqrt function returns the positive root only.
Table 3.1–1 Some common mathematical functions
Exponential
exp(x)
sqrt(x)
Logarithmic
log(x)
log10(x)
Complex
abs(x)
angle(x)
conj(x)
imag(x)
real(x)
Numeric
ceil(x)
fix(x)
floor(x)
round(x)
sign(x)
Exponential; ex.
Square root; 1x.
Natural logarithm; ln x.
Common (base-10) logarithm; log x log10 x.
Absolute value; x.
Angle of a complex number x.
Complex conjugate.
Imaginary part of a complex number x.
Real part of a complex number x.
Round to the nearest integer toward q .
Round to the nearest integer toward zero.
Round to the nearest integer toward - q .
Round toward the nearest integer.
Signum function:
1 if x 0; 0 if x 0; 1 if x 0.
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One of the strengths of MATLAB is that it will treat a variable as an array
automatically. For example, to compute the square roots of 5, 7, and 15, type
>>x = [5,7,15]; y = sqrt(x)
y =
2.2361
2.6358
3.8730
The square root function operates on every element in the array x.
Similarly, we can type exp(2) to obtain e2 7.3891, where e is the base of
the natural logarithms. Typing exp(1) gives 2.7183, which is e. Note that in
mathematics text, ln x denotes the natural logarithm, where x ey implies that
ln x = ln(ey) = y ln e = y
because ln e 1. However, this notation has not been carried over into MATLAB, which uses log(x) to represent ln x.
The common (base-10) logarithm is denoted in text by log x or log10 x. It is
de ned by the relation x 10y; that is,
log10 x = log10 10y = y log10 10 = y
because log1010 1. The MATLAB common logarithm function is log10(x).
A common mistake is to type log(x), instead of log10(x).
Another common error is to forget to use the array multiplication operator
.*. Note that in the MATLAB expression y = exp (x).*log(x), we need
to use the operator .* if x is an array because both exp(x) and log(x) will
be arrays.
Complex Number Functions
Chapter 1 explained how MATLAB easily handles complex number arithmetic.
In the rectangular representation the number a ib represents a point in the xy
plane. The number’s real part a is the x coordinate of the point, and the imaginary
part b is the y coordinate. The polar representation uses the distance M of the
point from the origin, which is the length of the hypotenuse, and the angle the
hypotenuse makes with the positive real axis. The pair (M, ) is simply the polar
coordinates of the point. From the Pythagorean theorem, the length of the
hypotenuse is given by M = 2a2 + b2 which is called the magnitude of the
number. The angle can be found from the trigonometry of the right triangle. It
is arctan (b/a).
Adding and subtracting complex numbers by hand is easy when they are in
the rectangular representation. However, the polar representation facilitates multiplication and division of complex numbers by hand. We must enter complex
numbers in MATLAB using the rectangular form, and its answers will be given
in that form. We can obtain the rectangular representation from the polar representation as follows:
a = M cos b = M sin 115
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The MATLAB abs(x) and angle(x) functions calculate the magnitude M and
angle of the complex number x. The functions real(x) and imag(x) return
the real and imaginary parts of x. The function conj(x) computes the complex
conjugate of x.
The magnitude of the product z of two complex numbers x and y is equal to
the product of their magnitudes: z = xy. The angle of the product is equal to
the sum of the angles: ∠ z = ∠ x + ∠ y. These facts are demonstrated below.
>>x = -3 + 4i; y = 6 - 8i;
>>mag_x = abs(x)
mag_x =
5.0000
>>mag_y = abs(y)
mag_y =
10.0000
>>mag_product = abs(x*y)
50.0000
>>angle_x = angle(x)
angle_x =
2.2143
>>angle_y = angle(y)
angle_y =
-0.9273
>>sum_angles = angle_x + angle_y
sum_angles =
1.2870
>>angle_product = angle(x*y)
angle_product =
1.2870
Similarly, for division, if z = x>y , then z = x>y and ∠ z = ∠x - ∠y.
Note that when x is a vector of real values, abs(x) does not give the geometric length of the vector. This length is given by norm(x). If x is a complex
number representing a geometric vector, then abs(x) gives its geometric length.
Numeric Functions
The round function rounds to the nearest integer. If y=[2.3,2.6,3.9],
typing round(y) gives the results 2, 3, 4. The x function truncates to the
nearest integer toward zero. Typing x(y) gives the results 2, 2, 3. The ceil
function (which stands for “ceiling”) rounds to the nearest integer toward q .
Typing ceil(y) produces the answers 3, 3, 4.
Suppose z = [-2.6,-2.3,5.7]. The oor function rounds to the
nearest integer toward - q . Typing oor(z) produces the result 3, 3, 5.
Typing x(z) produces the answer 2, 2, 5. The abs function computes the
absolute value. Thus abs(z) produces 2.6, 2.3, 5.7.
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Test Your Understanding
T3.1–1 For several values of x and y, con rm that ln ( xy) ⫽ ln x ⫹ ln y.
T3.1–2 Find the magnitude, angle, real part, and imaginary part of the
number 12 + 6i. (Answers: magnitude ⫽ 2.5149, angle ⫽ 0.6245 rad,
real part ⫽ 2.0402, imaginary part ⫽ 1.4705)
Function Arguments
When writing mathematics in text, we use parentheses ( ), brackets [ ], and
braces { } to improve the readability of expressions, and we have much latitude over their use. For example, we can write sin 2 in text, but MATLAB
requires parentheses surrounding the 2 (which is called the function argument
or parameter). Thus to evaluate sin 2 in MATLAB, we type sin(2). The
MATLAB function name must be followed by a pair of parentheses that surround the argument. To express in text the sine of the second element of the
array x, we would type sin[x(2)]. However, in MATLAB you cannot use
brackets or braces in this way, and you must type sin(x(2)).
You can include expressions and other functions as arguments. For example,
if x is an array, to evaluate sin(x2 ⫹ 5), you type sin(x.^2 + 5). To evaluate
sin( 1x + 1), you type sin(sqrt(x)+1). Be sure to check the order of
precedence and the number and placement of parentheses when typing such expressions. Every left-facing parenthesis requires a right-facing mate. However,
this condition does not guarantee that the expression is correct!
Another common mistake involves expressions such as sin2 x, which means
(sin x)2. In MATLAB, if x is a scalar we write this expression as (sin(x))^2, not
as sin^2(x), sin^2x, or sin(x^2)!
Trigonometric Functions
Other commonly used functions are cos(x), tan(x), sec(x), and csc(x),
which return cos x, tan x, sec x, and csc x, respectively. Table 3.1–2 lists the
MATLAB trigonometric functions that operate in radian mode. Thus sin(5)
computes the sine of 5 rad, not the sine of 5⬚. Similarly, the inverse trigonometric functions return an answer in radians. The functions that operate in degree
mode have the letter d appended to their names. For example, sind(x) accepts
the value of x in degrees. To compute the inverse sine in radians, type asin(x).
For example, asin(0.5) returns the answer 0.5236 rad. Note: In MATLAB,
sin(x)^(-1) does not give sin - 1(x) ; it gives 1/sin (x) !
MATLAB has two inverse tangent functions. The function atan(x) computes arctan x—the arctangent or inverse tangent—and returns an angle between
- ␲>2 and ␲>2. Another correct answer is the angle that lies in the opposite quadrant. The user must be able to choose the correct answer. For example, atan(1)
returns the answer 0.7854 rad, which corresponds to 45⬚. Thus tan 45⬚ ⫽ 1. However, tan(45⬚ ⫹ 180⬚) ⫽ tan 225⬚ ⫽ 1 also. Thus arctan(1) ⫽ 225⬚ is also correct.
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Table 3.1–2 Trigonometric functions
Trigonometric*
cos(x)
cot(x)
csc(x)
sec(x)
sin(x)
tan(x)
Inverse trigonometric†
acos(x)
acot(x)
acsc(x)
asec(x)
asin(x)
atan(x)
atan2(y,x)
Cosine; cos x.
Cotangent; cot x.
Cosecant; csc x.
Secant; sec x.
Sine; sin x.
Tangent; tan x.
Inverse cosine; arccos x cos1 x.
Inverse cotangent; arccot x cot1 x.
Inverse cosecant; arccsc x csc1 x.
Inverse secant; arcsec x sec1 x.
Inverse sine; arcsin x sin1 x.
Inverse tangent; arctan x tan1 x.
Four-quadrant inverse tangent.
*These functions accept x in radians.
†
These functions return a value in radians.
MATLAB provides the atan2 (y,x) function to determine the arctangent
unambiguously, where x and y are the coordinates of a point. The angle computed
by atan2(y,x) is the angle between the positive real axis and line from the
origin (0, 0) to the point (x, y). For example, the point x 1, y 1 corresponds
to 45 or 0.7854 rad, and the point x 1, y 1 corresponds to 135 or
2.3562 rad. Typing atan2 (-1,1) returns 0.7854, while typing
atan2 (1,-1) returns 2.3562. The atan2 (y,x) function is an example of a
function that has two arguments. The order of the arguments is important for such
functions. At present there is no atan2d function.
Test Your Understanding
T3.1–3 For several values of x, con rm that eix cos x i sin x.
T3.1–4 For several values of x in the range 0
cos1 x /2.
x
2 , con rm that sin 1 x T3.1–5 For several values of x in the range 0
2 tan x/(1 tan2 x).
x
2 , con rm that tan(2 x) Hyperbolic Functions
The hyperbolic functions are the solutions of some common problems in engineering analysis. For example, the catenary curve, which describes the shape of
a hanging cable supported at both ends, can be expressed in terms of the hyperbolic cosine, cosh x, which is de ned as
cosh x =
ex + e-x
2
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119
Table 3.1–3 Hyperbolic functions
Hyperbolic
cosh(x)
coth(x)
csch(x)
sech(x)
sinh(x)
tanh(x)
Inverse hyperbolic
acosh(x)
acoth(x)
acsch(x)
asech(x)
asinh(x)
atanh(x)
Hyperbolic cosine; cosh x (e x ex)/2.
Hyperbolic cotangent; cosh x /sinh x.
Hyperbolic cosecant; 1/sinh x.
Hyperbolic secant; 1/cosh x.
Hyperbolic sine; sinh x (e x ex)/2.
Hyperbolic tangent; sinh x/cosh x.
Inverse hyperbolic cosine
Inverse hyperbolic cotangent
Inverse hyperbolic cosecant
Inverse hyperbolic secant
Inverse hyperbolic sine
Inverse hyperbolic tangent
The hyperbolic sine, sinh x, is de ned as
ex - e - x
2
1
The inverse hyperbolic sine, sinh x, is the value y that satis es sinh y x.
Several other hyperbolic functions have been de ned. Table 3.1–3 lists these
hyperbolic functions and the MATLAB commands to obtain them.
sinh x =
Test Your Understanding
T3.1–6 For several values of x in the range 0
i sinh x.
T3.1–7 For several values of x in the range 10
sinh - 1 x = ln (x + 2x2 + 1).
5, con rm that sin( ix) x
x
10, con rm that
3.2 User-Defined Functions
Another type of M- le is a function le. Unlike a script le, all the variables in a
function le are local variables, which means their values are available only
within the function. Function les are useful when you need to repeat a set of
commands several times. They are the building blocks of larger programs.
To create a function le, open the Editor /Debugger as described in Chapter 1. The rst line in a function le must begin with a function de nition line that
has a list of inputs and outputs. This line distinguishes a function M- le from a
script M- le. Its syntax is as follows:
function [output variables] = function_name(input variables)
The output variables are those variables whose values are computed by the
function, using the given values of the input variables. Note that the output
FUNCTION FILE
LOCAL VARIABLE
FUNCTION
DEFINITION LINE
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variables are enclosed in square brackets (which are optional if there is only
one output), while the input variables must be enclosed with parentheses. The
function_name should be the same as the file name in which it is saved
(with the .m extension). That is, if we name a function drop, it should be
saved in the file drop.m. The function is “called” by typing its name (for example, drop) at the command line. The word function in the function definition line must be lowercase. Before naming a function, you can use the
exist function to see if another function has the same name.
Some Simple Function Examples
Functions operate on variables within their own workspace (called local variables), which is separate from the workspace you access at the MATLAB command prompt. Consider the following user-de ned function fun.
function z = fun(x,y)
u = 3*x;
z = u + 6*y.^2;
Note the use of the array exponentiation operator (.^). This enables the function to accept y as an array. Now consider what happens when you call this
function in various ways in the Command window. Call the function with its output argument:
>>x = 3; y = 7;
>>z = fun(x,y)
z =
303
or
>>z = fun(3,7)
z =
303
The function uses x 3 and y 7 to compute z.
Call the function without its output argument and try to access its value. You
see an error message.
>>clear z, fun(3,7)
ans =
303
>>z
???
Unde ned function or variable ‘z’.
Assign the output argument to another variable:
>>q = fun(3,7)
q =
303
You can suppress the output by putting a semicolon after the function call. For example, if you type q = fun(3,7); the value of q will be computed but not displayed.
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The variables x and y are local to the function fun, so unless you pass their
values by naming them x and y, their values will not be available in the
workspace outside the function. The variable u is also local to the function. For
example,
>>x = 3; y = 7; q = fun(x,y);
>>u
??? Unde ned function or variable ‘u’.
Compare this to
>>q = fun(3,7);
>>x
??? Unde ned function or variable ‘x’.
>>y
??? Unde ned function or variable ‘y’.
Only the order of the arguments is important, not the names of the arguments:
>>a = 7;b = 3;
>>z = fun(b,a)
z =
303
% This is equivalent to z = fun(3,7)
You can use arrays as input arguments:
>>r = fun([2:4],[7:9])
r =
300
393
498
A function may have more than one output. These are enclosed in square
brackets. For example, the function circle computes the area A and circumference C of a circle, given its radius as an input argument.
function [A, C] = circle(r)
A = pi*r.^2;
C = 2*pi*r;
The function is called as follows, if r 4.
>>[A, C] = circle(4)
A =
50.2655
C =
25.1327
A function may have no input arguments and no output list. For example, the
following user-de ned function show_date computes and stores the date in
the variable today, and displays the value of today.
function show_date
today = date
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Variations in the Function Line
The following examples show permissible variations in the format of the function line. The differences depend on whether there is no output, a single output,
or multiple outputs.
Function de nition line
File name
1. function [area_square] square(side);
2. function area_square square(side);
3. function volume_box box(height,width,length);
4. function [area_circle,circumf] circle(radius);
5. function sqplot(side);
square.m
square.m
box.m
circle.m
sqplot.m
Example 1 is a function with one input and one output. The square brackets are
optional when there is only one output (see example 2). Example 3 has one output
and three inputs. Example 4 has two outputs and one input. Example 5 has no output
variable (for example, a function that generates a plot). In such cases the equal sign
may be omitted.
Comment lines starting with the % sign can be placed anywhere in the function le. However , if you use help to obtain information about the function,
MATLAB displays all comment lines immediately following the function de nition line up to the rst blank line or rst executable line. The rst comment line
can be accessed by the lookfor command.
We can call both built-in and user-de ned functions either with the output
variables explicitly speci ed, as in examples 1 through 4, or without any output
variables speci ed. For example, we can call the function square as square
(side) if we are not interested in its output variable area_square. (The
function might perform some other operation that we want to occur, such as
producing a plot.) Note that if we omit the semicolon at the end of the function
call statement, the rst variable in the output variable list will be displayed using
the default variable name ans.
Variations in Function Calls
The following function, called drop, computes a falling object’s velocity and
distance dropped. The input variables are the acceleration g, the initial velocity y0, and the elapsed time t. Note that we must use the element-by-element
operations for any operations involving function inputs that are arrays. Here
we anticipate that t will be an array, so we use the element-by-element
operator (.^).
function [dist,vel] = drop(g,v0,t);
% Computes the distance traveled and the
% velocity of a dropped object, as functions
% of g, the initial velocity v0, and the time t.
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vel = g*t + v0;
dist = 0.5*g*t.^2 + v0*t;
The following examples show various ways to call the function drop:
1. The variable names used in the function de nition may , but need not, be
used when the function is called:
a = 32.2;
initial_speed = 10;
time = 5;
[feet_dropped,speed] = drop(a,initial_speed,time)
2. The input variables need not be assigned values outside the function prior
to the function call:
[feet_dropped,speed] = drop(32.2,10,5)
3. The inputs and outputs may be arrays:
[feet_dropped,speed]=drop(32.2,10,0:1:5)
This function call produces the arrays feet_dropped and speed, each with
six values corresponding to the six values of time in the array 0:1:5.
Local Variables
The names of the input variables given in the function de nition line are local to
that function. This means that other variable names can be used when you call the
function. All variables inside a function are erased after the function nishes executing, except when the same variable names appear in the output variable list
used in the function call.
For example, when using the drop function in a program, we can assign a
value to the variable dist before the function call, and its value will be unchanged after the call because its name was not used in the output list of the call
statement (the variable feet_dropped was used in the place of dist). This
is what is meant by the function’s variables being “local” to the function. This
feature allows us to write generally useful functions using variables of our
choice, without being concerned that the calling program uses the same variable
names for other calculations. This means that our function les are “portable”
and need not be rewritten every time they are used in a different program.
You might nd the M- le Debugger to be useful for locating errors in function les. Runtime errors in functions are more dif cult to locate because the
function’s local workspace is lost when the error forces a return to the MATLAB
base workspace. The Debugger provides access to the function workspace and
allows you to change values. It also enables you to execute lines one at a time and
to set breakpoints, which are speci c locations in the le where execution is temporarily halted. The applications in this text will probably not require use of the
Debugger, which is useful mainly for very large programs. For more information, see Chapter 4 of this text and also Chapter 4 of [Palm, 2005].
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Global Variables
The global command declares certain variables global, and therefore their values
are available to the basic workspace and to other functions that declare these variables global. The syntax to declare the variables A, X, and Q is global A X Q.
Use a space, not a comma, to separate the variables. Any assignment to those variables, in any function or in the base workspace, is available to all the other functions declaring them global. If the global variable doesn’t exist the rst time you
issue the global statement, it will be initialized to the empty matrix. If a variable
with the same name as the global variable already exists in the current workspace,
MATLAB issues a warning and changes the value of that variable to match the
global. In a user-de ned function, make the global command the rst executable
line. Place the same command in the calling program. It is customary, but not required, to capitalize the names of global variables and to use long names, to make
them easily recognizable.
The decision to declare a variable global is not always clear-cut. It is recommended to avoid using global variables. This can often be done by using
anonymous and nested functions, as discussed in Section 3.3.
Function Handles
A function handle is a way to reference a given function. First introduced in
MATLAB 6.0, function handles have become widely used and frequently appear in
examples throughout the MATLAB documentation. You can create a function handle to any function by using the @ sign before the function name. You can then give
the handle a name, if you wish, and you can use the handle to reference the function.
For example, consider the following user-de ned function, which computes
y x 2ex 3.
function y = f1(x)
y = x + 2*exp(-x) - 3;
To create a handle to this function and name the handle fh1, you type fh1 = @f1.
Function Functions
Some MATLAB functions act on functions. These commands are called function
functions. If the function acted upon is not a simple function, it is more convenient to de ne the function in an M- le. You can pass the function to the calling
function by using a function handle.
Finding the Zeros of a Function You can use the fzero function to nd the
zero of a function of a single variable, which is denoted by x. Its basic syntax is
fzero(@function, x0)
where @function is a function handle and x0 is a user-supplied guess for the
zero. The fzero function returns a value of x that is near x0. It identi es only
points where the function crosses the x axis, not points where the function just
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touches the axis. For example, fzero(@cos,2) returns the value x 1.5708.
As another example, y x2 is a parabola that touches the x axis at x 0. Because
the function never crosses the x axis, however, no zero will be found.
The function fzero(@function,x0) tries to nd a zero of function
near x0, if x0 is a scalar. The value returned by fzero is near a point where
function changes sign, or NaN if the search fails. In this case, the search terminates when the search interval is expanded until an Inf, NaN, or a complex value
is found (fzero cannot nd complex zeros). If x0 is a vector of length 2, fzero
assumes that x0 is an interval where the sign of function(x0(1)) differs from
the sign of function(x0(2)). An error occurs if this is not true. Calling fzero
with such an interval guarantees that fzero will return a value near a point where
function changes sign. Plotting the function rst is a good way to get a value for
the vector x0. If the function is not continuous, fzero might return values that
are discontinuous points instead of zeros. For example, x = fzero(@tan,1)
returns x = 1.5708, a discontinuous point in tan(x).
Functions can have more than one zero, so it helps to plot the function rst and
then use fzero to obtain an answer that is more accurate than the answer read
off the plot. Figure 3.2–1 shows the plot of the function y x 2ex 3, which
has two zeros, one near x 0.5 and one near x 3. Using the function le f1 created earlier to nd the zero near x 0.5, type x = fzero(@f1,-0.5). The
answer is x 0.5831. To nd the zero near x 3, type x = fzero(@f1,3).
The answer is x 2.8887.
2.5
2
1.5
y
1
0.5
0
–0.5
–1
–1.5
–1
0
1
2
x
Figure 3.2–1 Plot of the function y x 2ex 3.
3
4
5
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The syntax fzero (@f1,-0.5) is preferred to the older syntax fzero
(‘f1’, -0.5).
Minimizing a Function of One Variable The fminbnd function nds the
minimum of a function of a single variable, which is denoted by x. Its basic syntax is
fminbnd(@function, x1, x2)
where @function is a function handle. The fminbnd function returns a value
of x that minimizes the function in the interval x1
x
x2. For example,
fminbnd(@cos,0,4) returns the value x 3.1416.
However, to use this function to nd the minimum of more-complicated
functions, it is more convenient to de ne the function in a function le. For example, if y 1 xex, de ne the following function le:
function y = f2(x)
y = 1-x.*exp(-x);
To nd the value of x that gives a minimum of y for 0
x
5, type x =
fminbnd(@f2,0,5). The answer is x 1. To nd the minimum value of y,
type y = f2(x). The result is y = 0.6321.
Whenever we use a minimization technique, we should check that the
solution is a true minimum. For example, consider the polynomial y = 0.025x5 0.0625x4 - 0.333x3 + x2. Its plot is shown in Figure 3.2–2. The function has
two minimum points in the interval 1 x 4. The minimum near x 3 is
called a relative or local minimum because it forms a valley whose lowest point
is higher than the minimum at x 0. The minimum at x 0 is the true minimum
and is also called the global minimum. First create the function file
function y = f3(x)
y = polyva1([0.025, -0.0625, -0.333, 1, 0, 0], x);
To specify the interval 1 x 4, type x = fminbnd(@f3, -1, 4).
MATLAB gives the answer x = 2.0438e-006, which is essentially 0, the
true minimum point. If we specify the interval 0.1 x 2.5, MATLAB gives
the answer x = 0.1001, which corresponds to the minimum value of y on the
interval 0.1 x 2.5. Thus we will miss the true minimum point if our specied interval does not include it.
Also fminbnd can give misleading answers. If we specify the interval 1
x 4, MATLAB (R2009 b) gives the answer x = 2.8236, which corresponds
to the “valley” shown in the plot, but which is not the minimum point on the interval 1 x 4. On this interval the minimum point is at the boundary x 1. The
fminbnd procedure looks for a minimum point corresponding to a zero slope. In
practice, the best use of the fminbnd function is to determine precisely the
location of a minimum point whose approximate location was found by other
means, such as by plotting the function.
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4.5
4
3.5
3
y
2.5
2
1.5
1
0.5
0
–1
–0.5
0
0.5
1
1.5
x
2
2.5
3
3.5
4
Figure 3.2–2 Plot of the function y 0.025x 5 0.0625x 4 0.333x 3 x 2.
Minimizing a Function of Several Variables To nd the minimum of a function of more than one variable, use the fminsearch function. Its basic syntax is
fminsearch(@function, x0)
where @function is a function handle. The vector x0 is a guess that must be
2
2
supplied by the user. For example, to use the function f = xe-x - y , rst de ne it
in an M- le, using the vector x whose elements are x(1) = x and x(2) = y.
function f = f4(x)
f = x(1).*exp(-x(1).^2-x(2).^2);
Suppose we guess that the minimum is near x y 0. The session is
>>fminsearch(@f4, [0, 0])
ans =
-0.7071
0.000
Thus the minimum occurs at x 0.7071, y 0.
The fminsearch function can often handle discontinuities, particularly
if they do not occur near the solution. The fminsearch function might give
local solutions only, and it minimizes over the real numbers only; that is, x
must consist of real variables only, and the function must return real
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Table 3.2–1 Minimization and root- nding functions
Function
Description
fminbnd(@function,x1,x2)
Returns a value of x in the interval x1
x
x2 that corresponds to a minimum of the
single-variable function described by the
handle @function.
Uses the starting vector x0 to nd a minimum of the multivariable function described
by the handle @function.
Uses the starting value x0 to nd a zero of
the single-variable function described by the
handle @function.
fminsearch(@function,x0)
fzero(@function,x0)
numbers only. When x has complex values, they must be split into real and
imaginary parts.
Table 3.2–1 summarizes the basic syntax of the fminbnd, fminsearch,
and fzero commands.
These functions have extended syntax not described here. With these forms you
can specify the accuracy required for the solution as well as the number of steps to
use before stopping. Use the help facility to nd out more about these functions.
Optimization of an Irrigation Channel
EXAMPLE 3.2–1
Figure 3.2–3 shows the cross section of an irrigation channel. A preliminary analysis has
shown that the cross-sectional area of the channel should be 100 ft2 to carry the desired
water ow rate. To minimize the cost of concrete used to line the channel, we want to
minimize the length of the channel’s perimeter. Find the values of d, b, and that minimize this length.
■ Solution
The perimeter length L can be written in terms of the base b, depth d, and angle as follows:
L = b +
2d
sin The area of the trapezoidal cross section is
100 = db +
d2
tan d
θ
θ
b
Figure 3.2–3 Cross section of an irrigation channel.
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The variables to be selected are b, d, and . We can reduce the number of variables by
solving the latter equation for b to obtain
b =
1
d2
a100 b
d
tan Substitute this expression into the equation for L. The result is
100
d
2d
L =
+
d
tan sin We must now nd the values of d and to minimize L.
First de ne the function le for the perimeter length. Let the vector x be [d ].
function L = channel(x)
L = 100./x(1) - x(1)./tan(x(2)) + 2*x(1)./sin(x(2));
Then use the fminsearch function. Using a guess of d 20 and 1 rad, the session is
>>x = fminsearch (@channel,[20,1])
x =
7.5984
1.0472
Thus the minimum perimeter length is obtained with d 7.5984 ft and 1.0472 rad,
or 60 . Using a different guess, d 1, 0.1, produces the same answer. The value
of the base b corresponding to these values is b 8.7738.
However, using the guess d 20, 0.1 produces the physically meaningless
result d 781, 3.1416. The guess d 1, 1.5 produces the physically meaningless result d 3.6058, 3.1416.
The equation for L is a function of the two variables d and , and it forms a surface
when L is plotted versus d and on a three-dimensional coordinate system. This surface
might have multiple peaks, multiple valleys, and “mountain passes” called saddle points
that can fool a minimization technique. Different initial guesses for the solution vector
can cause the minimization technique to nd dif ferent valleys and thus report different results. We can use the surface-plotting functions covered in Chapter 5 to look for multiple
valleys, or we can use a large number of initial values for d and , say, over the physically
realistic ranges 0 d 30 and 0 /2. If all the physically meaningful answers
are identical, then we can be reasonably sure that we have found the minimum.
Test Your Understanding
T3.2–1 The equation e0.2x sin(x 2) 0.1 has three solutions in the interval
0 x 10. Find these three solutions. (Answers: x 1.0187, 4.5334,
7.0066)
T3.2–2 The function y 1 e0.2x sin(x 2) has two minimum points in
the interval 0 x 10. Find the values of x and y at each minimum.
(Answers: (x, y) (2.5150, 0.4070), (9.0001, 0.8347))
T3.2–3 Find the depth d and angle to minimize the perimeter length of the
channel shown in Figure 3.2–3 to provide an area of 200 ft2. (Answer:
d 10.7457 ft, 60 .)
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3.3 Additional Function Topics
In addition to function handles, anonymous functions, subfunctions, and nested
functions are some of the newer features of MATLAB. This section covers the
basic features of these new types of functions.
Methods for Calling Functions
There are four ways to invoke, or “call,” a function into action:
1.
2.
3.
4.
As a character string identifying the appropriate function M- le
As a function handle
As an “inline” function object
As a string expression
Examples of these ways follow for the fzero function used with the userde ned function fun1, which computes y x2 4.
1. As a character string identifying the appropriate function M- le, which is
function y = fun1(x)
y = x.^2-4;
The function may be called as follows, to compute the zero over the range
0 x 3:
>>x = fzero(‘fun1’,[0, 3])
2. As a function handle to an existing function M- le:
>>x = fzero(@fun1,[0, 3])
3. As an “inline” function object:
>>fun1 = ‘x.^2-4’;
>>fun_inline = inline(fun1);
>>x = fzero(fun_inline,[0, 3])
4. As a string expression:
>>fun1 = ‘x.^2-4’;
>>x = fzero(fun1,[0, 3])
or as
>>x = fzero(‘x.^2-4’,[0, 3])
Method 2 was not available prior to MATLAB 6.0, and it is now preferred
over method 1. The third method is not discussed in this text because it is a
slower method than the rst two. The third and fourth methods are equivalent because they both utilize the inline function; the only difference is that with the
fourth method MATLAB determines that the rst ar gument of fzero is a string
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131
variable and calls inline to convert the string variable to an inline function object. The function handle method (method 2) is the fastest method, followed by
method 1.
In addition to speed improvement, another advantage of using a function
handle is that it provides access to subfunctions, which are normally not visible
outside of their de ning M- le. This is discussed later in this section.
Types of Functions
At this point it is helpful to review the types of functions provided for in MATLAB. MATLAB provides built-in functions, such as clear, sin, and plot,
which are not M- les, and some functions that are M- les, such as the function
mean. In addition, the following types of user-de ned functions can be created
in MATLAB.
■
■
■
■
■
■
The primary function is the rst function in an M- le and typically contains the main program. Following the primary function in the same le
can be any number of subfunctions, which can serve as subroutines to the
primary function. Usually the primary function is the only function in an
M- le that you can call from the MA TLAB command line or from another
M- le function. You invoke this function by using the name of the M- le
in which it is de ned. We normally use the same name for the function and
its le, but if the function name dif fers from the le name, you must use the
le name to invoke the function.
Anonymous functions enable you to create a simple function without needing to create an M- le for it. You can construct an anonymous function
either at the MATLAB command line or from within another function or
script. Thus, anonymous functions provide a quick way of making a function from any MATLAB expression without the need to create, name, and
save a le.
Subfunctions are placed in the primary function and are called by the primary function. You can use multiple functions within a single primary
function M- le.
Nested functions are functions de ned within another function. They can
help to improve the readability of your program and also give you more
exible access to variables in the M- le. The difference between nested
functions and subfunctions is that subfunctions normally cannot be
accessed outside of their primary function le.
Overloaded functions are functions that respond differently to different
types of input arguments. They are similar to overloaded functions in any
object-oriented language. For example, an overloaded function can be
created to treat integer inputs differently than inputs of class double.
Private functions enable you to restrict access to a function. They can be
called only from an M- le function in the parent directory .
PRIMARY
FUNCTION
ANONYMOUS
FUNCTIONS
SUBFUNCTIONS
NESTED
FUNCTIONS
PRIVATE
FUNCTION
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Anonymous Functions
Anonymous functions enable you to create a simple function without needing to
create an M- le for it. You can construct an anonymous function either at the
MATLAB command line or from within another function or script. The syntax
for creating an anonymous function from an expression is
fhandle = @(arglist) expr
where arglist is a comma-separated list of input arguments to be passed to the
function and expr is any single, valid MATLAB expression. This syntax creates
the function handle fhandle, which enables you to invoke the function. Note
that this syntax is different from that used to create other function handles,
fhandle = @functionname. The handle is also useful for passing the
anonymous function in a call to some other function in the same way as any other
function handle.
For example, to create a simple function called sq to calculate the square of
a number, type
sq = @(x) x.^2;
To improve readability, you may enclose the expression in parentheses, as sq =
@(x) (x.^2);. To execute the function, type the name of the function handle,
followed by any input arguments enclosed in parentheses. For example,
>>sq(5)
ans =
25
>>sq([5,7])
ans =
25
49
You might think that this particular anonymous function will not save you any
work because typing sq([5,7]) requires nine keystrokes, one more than is required to type [5,7].^2. Here, however, the anonymous function protects you
from forgetting to type the period (.) required for array exponentiation. Anonymous functions are useful, however, for more complicated functions involving
numerous keystrokes.
You can pass the handle of an anonymous function to other functions. For
example, to nd the minimum of the polynomial 4 x2 50x 5 over the interval [10, 10], you type
>>poly1 = @(x) 4*x.^2 - 50*x + 5;
>>fminbnd(poly1, -10, 10)
ans =
6.2500
If you are not going to use that polynomial again, you can omit the handle de nition line and type instead
>>fminbnd(@(x) 4*x.^2 - 50*x + 5, -10, 10)
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Multiple-Input Arguments You can create anonymous functions having more
than one input. For example, to de ne the function 2x2 + y2, type
>>sqrtsum = @(x,y) sqrt(x.^2 + y.^2);
Then
>>sqrtsum(3, 4)
ans =
5
As another example, consider the function z Ax By de ning a plane.
The scalar variables A and B must be assigned values before you create the function handle. For example,
>>A = 6; B = 4:
>>plane = @(x,y) A*x + B*y;
>>z = plane(2,8)
z =
44
No-Input Arguments To construct a handle for an anonymous function that
has no input arguments, use empty parentheses for the input argument list, as
shown by the following: d = @() date;.
Use empty parentheses when invoking the function, as follows:
>>d()
ans =
01-Mar-2007
You must include the parentheses. If you do not, MATLAB just identi es the
handle; it does not execute the function.
Calling One Function within Another One anonymous function can call another to implement function composition. Consider the function 5 sin(x3). It is
composed of the functions g(y) 5 sin(y) and f(x) x3. In the following session
the function whose handle is h calls the functions whose handles are f and g to
compute 5 sin (23).
>>f = @(x) x.^3;
>>g = @(x) 5*sin(x);
>>h = @(x) g(f(x));
>>h(2)
ans =
4.9468
To preserve an anonymous function from one MATLAB session to the next,
save the function handle to a MAT- le. For example, to save the function
associated with the handle h, type save anon.mat h. To recover it in a later
session, type load anon.mat h.
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Variables and Anonymous Functions Variables can appear in anonymous
functions in two ways:
■ As variables speci ed in the ar gument list, such as f = @(x) x.^3;.
■ As variables speci ed in the body of the expression, such as with the
variables A and B in plane = @(x,y) A*x + B*y. In this case,
when the function is created, MATLAB captures the values of these variables and retains those values for the lifetime of the function handle. In this
example, if the values of A or B are changed after the handle is created,
their values associated with the handle do not change. This feature has both
advantages and disadvantages, so you must keep it in mind.
Subfunctions
A function M- le may contain more than one user -de ned function. The rst
de ned function in the le is called the primary function, whose name is the
same as the M- le name. All other functions in the le are called subfunctions.
Subfunctions are normally “visible” only to the primary function and to other
subfunctions in the same le; that is, they normally cannot be called by programs
or functions outside the le. However , this limitation can be removed with the
use of function handles, as we will see later in this section.
Create the primary function rst with a function de nition line and its de ning
code, and name the le with this function name as usual. Then create each subfunction with its own function de nition line and de ning code. The order of the subfunctions does not matter, but function names must be unique within the M- le.
The order in which MATLAB checks for functions is very important. When a
function is called from within an M- le, MA TLAB rst checks to see if the function
is a built-in function such as sin. If not, it checks to see if it is a subfunction in the
le, then checks to see if it is a private function (which is a function M- le residing
in the private subdirectory of the calling function). Then MATLAB checks for a
standard M- le on your search path. Thus, because MATLAB checks for a subfunction before checking for private and standard M- le functions, you may use subfunctions with the same name as another existing M- le. This feature allows you to
name subfunctions without being concerned about whether another function exists
with the same name, so you need not choose long function names to avoid con ict.
This feature also protects you from using another function unintentionally.
Note that you may even supercede a MATLAB M-function in this way. The
following example shows how the MATLAB M-function mean can be superceded
by our own de nition of the mean, one which gives the root-mean-square value.
The function mean is a subfunction. The function subfun_demo is the primary
function.
function y = subfun_demo(a)
y = a - mean(a);
%
function w = mean(x)
w = sqrt(sum(x.^2))/length(x);
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A sample session follows.
>>y = subfn_demo([4, -4])
y =
1.1716
-6.8284
If we had used the MATLAB M-function mean, we would have obtained a
different answer, that is,
>>a=[4,-4];
>>b = a - mean(a)
b =
4
-4
Thus the use of subfunctions enables you to reduce the number of les that
de ne your functions. For example, if it were not for the subfunction mean in the
previous example, we would have had to de ne a separate M- le for our mean
function and give it a different name so as not to confuse it with the MATLAB
function of the same name.
Subfunctions are normally visible only to the primary function and other
subfunctions in the same le. However , we can use a function handle to allow access to the subfunction from outside the M- le, as the following example shows.
Create the following M- le with the primary function fn_demo1(range)
and the subfunction testfun(x) to compute the zeros of the function
(x2 4) cos x over the range speci ed in the input variable range. Note the use
of a function handle in the second line.
function yzero = fn_demo1(range)
fun = @testfun;
[yzero,value] = fzero(fun,range);
%
function y = testfun(x)
y = (x.^2-4).*cos(x);
A test session gives the following results.
>>yzero = fn_demo1([3, 6])
yzero =
4.7124
So the zero of (x2 4) cos x over 3
x
6 occurs at x 4.7124.
Nested Functions
With MATLAB 7 you can now place the de nitions of one or more functions
within another function. Functions so de ned are said to be nested within the
main function. You can also nest functions within other nested functions. Like
any M- le function, a nested function contains the usual components of an M- le
function. You must, however, always terminate a nested function with an end
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statement. In fact, if an M- le contains at least one nested function, you must terminate all functions, including subfunctions, in the le with an end statement,
whether or not they contain nested functions.
The following example constructs a function handle for a nested function
p(x) and then passes the handle to the MATLAB function fminbnd to nd
the minimum point on the parabola. The parabola function constructs and returns a function handle f for the nested function p that evaluates the parabola
ax2 bx c. This handle gets passed to fminbnd.
function f = parabola(a, b, c)
f = @p;
% Nested function
function y = p(x)
y = polyval ([a,b,c],x);
end
end
In the Command window type
>>f = parabola(4, -50, 5);
>>fminbnd(f, -10, 10)
ans =
6.2500
Note than the function p(x) can see the variables a, b, and c in the calling
function’s workspace.
Contrast this approach to that required using global variables. First create the
function p(x).
function y = p(x)
global a b c
y = polyval ([a, b, c], x);
Then, in the Command window, type
>>global a b c
>>a = 4; b = -50; c = 5;
>> fminbnd (@p, -10, 10)
Nested functions might seem to be the same as subfunctions, but they are
not. Nested functions have two unique properties:
1. A nested function can access the workspaces of all functions inside of
which it is nested. So, for example, a variable that has a value assigned to it
by the primary function can be read or overwritten by a function nested at
any level within the main function. In addition, a variable assigned in a
nested function can be read or overwritten by any of the functions containing that function.
2. If you construct a function handle for a nested function, the handle not only
stores the information needed to access the nested function, but also stores
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the values of all variables shared between the nested function and those
functions that contain it. This means that these variables persist in memory
between calls made by means of the function handle.
Consider the following representation of some functions named A, B, . . ., E.
function A(x, y)
B(x, y);
D(y);
function B(x, y)
C(x);
D(y);
% The primary function
% Nested in A
function C(x)
% Nested in B
D(x);
end
% This terminates C
end
% This terminates B
function D(x)
E(x);
end
% Nested in A
function E
% Nested in D
...
end
% This terminates E
end
% This terminates D
% This terminates A
You call a nested function in several ways.
1. You can call it from the level immediately above it. (In the previous code,
function A can call B or D, but not C or E.)
2. You can call it from a function nested at the same level within the same
parent function. (Function B can call D, and D can call B.)
3. You can call it from a function at any lower level. (Function C can call
B or D, but not E.)
4. If you construct a function handle for a nested function, you can call the
nested function from any MATLAB function that has access to the handle.
You can call a subfunction from any nested function in the same M- le.
Private Functions
Private functions reside in subdirectories with the special name private, and
they are visible only to functions in the parent directory. Assume the directory
rsmith is on the MATLAB search path. A subdirectory of rsmith called private may contain functions that only the functions in rsmith can call. Because
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private functions are invisible outside the parent directory rsmith, they can use
the same names as functions in other directories. This is useful if the main
directory is used by several individuals including R. Smith, but R. Smith wants to
create a personal version of a particular function while retaining the original in
the main directory. Because MATLAB looks for private functions before standard
M- le functions, it will nd a private function named, say , cylinder.m before a
nonprivate M- le named cylinder.m.
Primary functions and subfunctions can be implemented as private functions. Create a private directory by creating a subdirectory called private
using the standard procedure for creating a directory or a folder on your computer, but do not place the private directory on your path.
3.4 Working with Data Files
An ASCII data le may have one or more lines of text, called the header , at the
beginning. These might be comments that describe what the data represent, the
date they were created, and who created the data, for example. One or more lines
of data, arranged in rows and columns, follow the header. The numbers in each
row might be separated by spaces or by commas.
If it is inconvenient to edit the data le, the MA TLAB environment provides
many ways to bring data created by other applications into the MATLAB
workspace, a process called importing data, and to package workspace variables
so that they can be exported to other applications.
If the le has a header or the data are separated by commas, MA TLAB will
produce an error message. To correct this situation, rst load the data le into a
text editor, remove the header, and replace the commas with spaces. To retrieve
these data into MATLAB, type load filename. If the le has m lines with n
values in each line, the data will be assigned to an m n matrix having the same
name as the le with the extension stripped of f. Your data le can have any extension except .mat, so that MATLAB will not try to load the le as a
workspace le.
Importing Spreadsheet Files
Some spreadsheet programs store data in the .wk1 format. You can use the command M = wk1read(‘filename’) to import these data into MATLAB and
store them in the matrix M. The command A = xlsread(‘filename’)
imports the Microsoft Excel workbook le filename.xls into the array A.
The command [A, B] = xlsread(‘filename’) imports all numeric
data into the array A and all text data into the cell array B.
The Import Wizard
You can use the Import Wizard to import many types of ASCII data formats, including data on the clipboard. The Import Wizard presents a series of dialog
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boxes in which you specify the name of the le, the delimiter used in the le, and
the variables that you want to import.
Do the following to import this sample tab-delimited, ASCII data le,
testdata.txt:
1
17
2
12
3
8
4
15
5;
25;
1. Activate the Import Wizard either by typing uiimport or by selecting the
Import Data option on the MATLAB Desktop File menu. The Import
Wizard displays a dialog box that asks you to specify the name of the le
you want to import.
2. The Import Wizard processes the contents of the le and displays tabs identifying the variables it recognizes in the le, and displays a portion of the
data in a grid, similar to a spreadsheet. The Import Wizard uses the space
character as the default delimiter. After you click Next, the Import Wizard
attempts to identify the delimiter (see Figure 3.4–1).
3. In the next dialog box, the Import Wizard displays a list of the variables it
found in the le. It also displays the contents of the rst variable in the list.
In this example there is only one variable, named testdata.
4. Choose the variables you want to import by clicking the check boxes next
to their names. By default, all variables are checked for import. After you
select the variables you want to import, click the Finish button to import
the data into the MATLAB workspace.
To import data from the clipboard, select Paste Special from the Edit menu.
Then proceed with step 2. The default variable name is A_pastespecial.
Exporting ASCII Data Files
You might want to export a MATLAB matrix as an ASCII data le where the
rows and columns are represented as space-delimited, numeric values. To export
a MATLAB matrix as a delimited ASCII data le, you can use either the save
Figure 3.4–1 The rst screen in the Import Wizard.
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command, specifying the -ASCII quali er , or the dlmwrite function. The
save command is easy to use; however, the dlmwrite function provides
greater exibility , allowing you to specify any character as a delimiter and to export subsets of an array by specifying a range of values.
Suppose you have created the array A = [1 2 3 4; 5 6 7 8] in
MATLAB. To export the array using the save command, type the following in
the Command window.
>>save my_data.out A -ASCII
By default, save uses spaces as delimiters, but you can use tabs instead of
spaces by specifying the -tab quali er .
3.5 Summary
In Section 3.1 we introduced just some of the most commonly used mathematical functions. You should now be able to use the MATLAB Help to nd other
functions you need. If necessary, you can create your own functions, using the
methods of Section 3.2. This section also covered function handles and their use
with function functions.
Anonymous functions, subfunctions, and nested functions extend the capabilities of MATLAB. These topics were treated in Section 3.3. In addition to
function les, data les are useful for many applications. Section 3.4 shows how
to import and export such les in MA TLAB.
Key Terms with Page References
Anonymous functions, 131
Function argument, 117
Function de nition line, 119
Function le, 119
Function handle, 124
Global variables, 124
Local variable, 119
Nested functions, 131
Primary function, 131
Private function, 131
Subfunctions, 131
Problems
You can nd the answers to problems marked with an asterisk at the end of the text.
Section 3.1
1.*
Suppose that y 3 ix. For x 0, 1, and 2, use MATLAB to compute
the following expressions. Hand-check the answers.
a. y
b. 1y
c. (5 7i)y
y
d. 6 - 3i
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Problems
2.*
Let x 5 8i and y 10 5i. Use MATLAB to compute the following
expressions. Hand-check the answers.
a. The magnitude and angle of xy.
b. The magnitude and angle of xy .
3.*
Use MATLAB to nd the angles corresponding to the following coordinates. Hand-check the answers.
a. (x, y) (5, 8)
b. (x, y) (5, 8)
c. (x, y) (5, 8)
d. (x, y) (5, 8)
4.
For several values of x, use MATLAB to con rm that sinh x (e x ex)/2.
5.
For several values of x, use MATLAB to con rm that cosh-1 x =
ln (x + 1x2 - 1), x Ú 1
6.
The capacitance of two parallel conductors of length L and radius r, separated by a distance d in air, is given by
C =
L
ln [(d - r)/r]
where is the permittivity of air ( 8.854 1012 F/m).
Write a script le that accepts user input for d, L, and r and computes
and displays C. Test the le with the values L 1 m, r 0.001 m, and
d 0.004 m.
7.*
When a belt is wrapped around a cylinder, the relation between the belt
forces on each side of the cylinder is
F1 = F2 e
where is the angle of wrap of the belt and is the friction coef cient.
Write a script le that rst prompts a user to specify , , and F2 and
then computes the force F1. Test your program with the values 130,
0.3, and F2 100 N. (Hint: Be careful with !)
Section 3.2
8.
The output of the MATLAB atan2 function is in radians. Write a function called atan2d that produces an output in degrees.
9.
Write a function that accepts temperature in degrees Fahrenheit (°F) and
computes the corresponding value in degrees Celsius (°C). The relation
between the two is
T °C =
Be sure to test your function.
5
(T °F - 32)
9
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10.* An object thrown vertically with a speed y0 reaches a height h at time t,
where
h = y0t -
1 2
gt
2
Write and test a function that computes the time t required to reach a
speci ed height h, for a given value of y0. The function’s inputs should be
h, y0, and g. Test your function for the case where h 100 m, y0 50 m/s,
and g 9.81 m/s2. Interpret both answers.
11.
A water tank consists of a cylindrical part of radius r and height h and a
hemispherical top. The tank is to be constructed to hold 600 m3 when
lled. The surface area of the cylindrical part is 2 rh, and its volume is
r2h. The surface area of the hemispherical top is given by 2 r2, and its
volume is given by 2 r3/3. The cost to construct the cylindrical part of
the tank is $400 per square meter of surface area; the hemispherical part
costs $600 per square meter. Use the fminbnd function to compute the
radius that results in the least cost. Compute the corresponding height h.
12.
A fence around a eld is shaped as shown in Figure P12. It consists of a
rectangle of length L and width W, and a right triangle that is
symmetrical about the central horizontal axis of the rectangle. Suppose
the width W is known (in meters) and the enclosed area A is known
(in square meters). Write a user-de ned function le with W and A as
inputs. The outputs are the length L required so that the enclosed area is A
and the total length of fence required. Test your function for the
values W 6 m and A 80 m2.
L
D
W
Figure P12
13.
A fenced enclosure consists of a rectangle of length L and width 2R and a
semicircle of radius R, as shown in Figure P13. The enclosure is to be
built to have an area A of 2000 ft2. The cost of the fence is $50 per foot
for the curved portion and $40 per foot for the straight sides. Use the
fminbnd function to determine with a resolution of 0.01 ft the values of
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Problems
R and L required to minimize the total cost of the fence. Also compute the
minimum cost.
L
R
2R
Figure P13
14.
Using estimates of rainfall, evaporation, and water consumption, the town
engineer developed the following model of the water volume in the reservoir as a function of time
V(t) = 109 + 108(1 - e-t>100) - rt
where V is the water volume in liters, t is time in days, and r is the town’s
consumption rate in liters per day. Write two user-de ned functions. The
rst function should de ne the function V(t) for use with the fzero function. The second function should use fzero to compute how long it will
take for the water volume to decrease to x percent of its initial value of
109 L. The inputs to the second function should be x and r. Test your
functions for the case where x 50 percent and r 107 L/day.
15.
The volume V and paper surface area A of a conical paper cup are given by
V =
1 2
rh
3
A =
r 2r2 + h2
where r is the radius of the base of the cone and h is the height of the
cone.
a. By eliminating h, obtain the expression for A as a function of r and V.
b. Create a user-de ned function that accepts R as the only argument and
computes A for a given value of V. Declare V to be global within the
function.
c. For V 10 in.3, use the function with the fminbnd function to compute the value of r that minimizes the area A. What is the corresponding value of the height h? Investigate the sensitivity of the solution by
plotting V versus r. How much can R vary about its optimal value before the area increases 10 percent above its minimum value?
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A torus is shaped like a doughnut. If its inner radius is a and its outer radius
is b, its volume and surface area are given by
V =
1
4
2
(a + b)(b - a)2
A =
2
(b2 - a2)
a. Create a user-de ned function that computes V and A from the arguments a and b.
b. Suppose that the outer radius is constrained to be 2 in. greater than the
inner radius. Write a script le that uses your function to plot A and V
versus a for 0.25 a 4 in.
17.
Suppose it is known that the graph of the function y ax3 bx2 cx d
passes through four given points (xi , yi ), i 1, 2, 3, 4. Write a userde ned function that accepts these four points as input and computes the
coef cients a, b, c, and d. The function should solve four linear equations
in terms of the four unknowns a, b, c, and d. Test your function for the
case where (xi , yi ) (2, 20), (0, 4), (2, 68), and (4, 508), whose
answer is a 7, b 5, c 6, and d 4.
Section 3.3
18.
Create an anonymous function for 10e2x and use it to plot the function
over the range 0 x 2.
19.
Create an anonymous function for 20x2 200x 3 and use it
a. To plot the function to determine the approximate location of its
minimum
b. With the fminbnd function to precisely determine the location of the
minimum
20.
Create four anonymous functions to represent the function 6e3 cos x ,
which is composed of the functions h(z) 6ez, g(y) 3 cos y, and
2
f (x) x2. Use the anonymous functions to plot 6e3 cos x over the range
0 x 4.
21.
Use a primary function with a subfunction to compute the zeros of the
function 3x3 12x2 33x 80 over the range 10 x 10.
22.
Create a primary function that uses a function handle with a nested function to compute the minimum of the function 20x2 200x 12 over the
range 0 x 10.
2
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Problems
Section 3.4
23.
Use a text editor to create a le containing the following data. Then use
the load function to load the le into MA TLAB, and use the mean function to compute the mean value of each column.
55
42
98
49
39
95
63
51
92
58
45
90
24.
Enter and save the data given in Problem 23 in a spreadsheet. Then import the spreadsheet le into the MA TLAB variable A. Use MATLAB to
compute the sum of each column.
25.
Use a text editor to create a le from the data given in Problem 23, but
separate each number with a semicolon. Then use the Import Wizard to
load and save the data in the MATLAB variable A.
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Courtesy Henry Guckel (Late)/Dept. of Electrical
Engineering, University of Wisconsin.
Engineering in the
21st Century. . .
Nanotechnology
W
hile large-scale technology is attracting much public attention, many
of the engineering challenges and opportunities in the 21st century
will involve the development of extremely small devices and even
the manipulation of individual atoms. This technology is called nanotechnology
because it involves processing materials whose size is about 1 nanometer (nm),
which is 109 m, or 1兾1 000 000 mm. The distance between atoms in singlecrystal silicon is 0.5 nm.
Nanotechnology is in its infancy, although some working devices have been
created. The micromotor with gear train shown above has a dimension of approximately 104 m. This device converts electrical input power into mechanical
motion. It was constructed using the magnetic properties of electroplated metal
lms.
While we are learning how to make such devices, another challenge is to develop innovative applications for them. Many of the applications proposed thus
far are medical; small pumps for drug delivery and surgical tools are two
examples. Researchers at the Lawrence Livermore Laboratory have developed a
microgripper tool to treat brain aneurysms. It is about the size of a grain of sand
and was constructed from silicon cantilever beams powered by a shape-memory
alloy actuator. To design and apply these devices, engineers must rst model the
appropriate mechanical and electrical properties. The features of MATLAB
provide excellent support for such analyses. ■
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C H A P T E R
4
Programming
with MATLAB
OUTLINE
4.1 Program Design and Development
4.2 Relational Operators and Logical Variables
4.3 Logical Operators and Functions
4.4 Conditional Statements
4.5 for Loops
4.6 while Loops
4.7 The switch Structure
4.8 Debugging MATLAB Programs
4.9 Applications to Simulation
4.10 Summary
Problems
The MATLAB interactive mode is very useful for simple problems, but morecomplex problems require a script le. Such a le can be called a computer program, and writing such a le is called programming. Section 4.1 presents a general
and ef cient approach to the design and development of programs.
The usefulness of MATLAB is greatly increased by the use of decisionmaking functions in its programs. These functions enable you to write programs
whose operations depend on the results of calculations made by the program.
Sections 4.2, 4.3, and 4.4 deal with these decision-making functions.
MATLAB can also repeat calculations a speci ed number of times or until
some condition is satis ed. This feature enables engineers to solve problems of
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great complexity or requiring numerous calculations. These “loop” structures are
covered in Sections 4.5 and 4.6.
The switch structure enhances the MATLAB decision-making capabilities. This topic is covered in Section 4.7. Use of the MATLAB Editor/Debugger
for debugging programs is covered in Section 4.8.
Section 4.9 discusses “simulation,” a major application of MATLAB programs that enables us to study the operation of complicated systems, processes,
and organizations. Tables summarizing the MATLAB commands introduced in
this chapter appear throughout the chapter, and Table 4.10–1 will help you locate
the information you need.
4.1 Program Design and Development
In this chapter we introduce relational operators, such as > and ==, and the two
types of loops used in MATLAB, the for loop and the while loop. These features, plus MATLAB functions and the logical operators to be introduced in
Section 4.3, form the basis for constructing MATLAB programs to solve complex problems. Design of computer programs to solve complex problems needs
to be done in a systematic manner from the start to avoid time-consuming and
frustrating dif culties later in the process. In this section we show how to structure and manage the program design process.
Algorithms and Control Structures
An algorithm is an ordered sequence of precisely de ned instructions that performs some task in a nite amount of time. An ordered sequence means that the
instructions can be numbered, but an algorithm often must have the ability to
alter the order of its instructions using what is called a control structure. There
are three categories of algorithmic operations:
Sequential operations. These instructions are executed in order.
Conditional operations. These control structures rst ask a question to be
answered with a true/false answer and then select the next instruction based
on the answer.
Iterative operations (loops). These control structures repeat the execution
of a block of instructions.
Not every problem can be solved with an algorithm, and some potential algorithmic solutions can fail because they take too long to nd a solution.
Structured Programming
Structured programming is a technique for designing programs in which a hierarchy of modules is used, each having a single entry and a single exit point, and
in which control is passed downward through the structure without unconditional
branches to higher levels of the structure. In MATLAB these modules can be
built-in or user-de ned functions.
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Program Design and Development
Control of the program ow uses the same three types of control structures
used in algorithms: sequential, conditional, and iterative. In general, any computer program can be written with these three structures. This realization led to
the development of structured programming. Languages suitable for structured
programming, such as MATLAB, thus do not have an equivalent to the goto
statement that you might have seen in the BASIC and FORTRAN languages. An
unfortunate result of the goto statement was confusing code, called spaghetti
code, composed of a complex tangle of branches.
Structured programming, if used properly, results in programs that are easy
to write, understand, and modify. The advantages of structured programming are
as follows.
1. Structured programs are easier to write because the programmer can study
the overall problem rst and deal with the details later .
2. Modules (functions) written for one application can be used for other applications (this is called reusable code).
3. Structured programs are easier to debug because each module is designed
to perform just one task, and thus it can be tested separately from the other
modules.
4. Structured programming is effective in a teamwork environment because
several people can work on a common program, each person developing
one or more modules.
5. Structured programs are easier to understand and modify, especially if
meaningful names are chosen for the modules and if the documentation
clearly identi es the module’ s task.
Top-down Design and Program Documentation
A method for creating structured programs is top-down design, which aims to
describe a program’s intended purpose at a very high level initially and then
partition the problem repeatedly into more detailed levels, one level at a time,
until enough is understood about the program structure to enable it to be coded.
Table 4.1–1, which is repeated from Chapter 1, summarizes the process of
Table 4.1–1 Steps for developing a computer solution
1.
2.
3.
4.
5.
6.
7.
8.
State the problem concisely.
Specify the data to be used by the program. This is the input.
Specify the information to be generated by the program. This is the output.
Work through the solution steps by hand or with a calculator; use a simpler set of data if
necessary.
Write and run the program.
Check the output of the program with your hand solution.
Run the program with your input data and perform a “reality check” on the output. Does
it make sense? Estimate the range of the expected result and compare it with you answer.
If you will use the program as a general tool in the future, test it by running it for a
range of reasonable data values; perform a reality check on the results.
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150
STRUCTURE
CHART
FLOWCHART
top-down design. In step 4 you create the algorithms used to obtain the solution.
Note that step 5, Write and run the program, is only part of the top-down design
process. In this step you create the necessary modules and test them separately.
Two types of charts aid in developing structured programs and in documenting them. These are structure charts and owcharts . A structure chart is a graphical description showing how the different parts of the program are connected. This
type of diagram is particularly useful in the initial stages of top-down design.
A structure chart displays the organization of a program without showing the
details of the calculations and decision processes. For example, we can create
program modules using function les that do speci c, readily identi able tasks.
Larger programs are usually composed of a main program that calls on the modules to do their specialized tasks as needed. A structure chart shows the connection between the main program and the modules.
For example, suppose you want to write a program that plays a game, say,
Tic-Tac-Toe. You need a module to allow the human player to input a move, a
module to update and display the game grid, and a module that contains the computer’s strategy for selecting its moves. Figure 4.1–1 shows the structure chart of
such a program.
Flowcharts are useful for developing and documenting programs that contain conditional statements, because they can display the various paths (called
branches) that a program can take, depending on how the conditional statements
are executed. The owchart representation of the verbal description of the if
statement (covered in Section 4.3) is shown in Figure 4.1–2. Flowcharts use the
diamond symbol to indicate decision points.
The usefulness of structure charts and owcharts is limited by their size. For
large, more complicated programs, it might be impractical to draw such charts.
Nevertheless, for smaller projects, sketching a owchart and/or a structure chart
might help you organize your thoughts before you begin to write the speci c
MATLAB code. Because of the space required for such charts we do not use them
in this text. You are encouraged, however, to use them when solving problems.
Main Program
Player Input
Program
Game Status
Display Program
Figure 4.1–1 Structure chart of a game program.
Strategy
Program
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Program Design and Development
Start
Logical
Expression
False
True
Statements
End
Figure 4.1–2 Flowchart
representation of the verbal
description of the if statement.
Documenting programs properly is very important, even if you never give
your programs to other people. If you need to modify one of your programs, you
will nd that it is often very dif cult to recall how it operates if you have not used
it for some time. Effective documentation can be accomplished with the use of
1.
2.
3.
4.
5.
Proper selection of variable names to re ect the quantities they represent.
Comments within the program.
Structure charts.
Flowcharts.
A verbal description of the program, often in pseudocode.
The advantage of using suitable variable names and comments is that they reside
with the program; anyone who gets a copy of the program will see such documentation. However, they often do not provide enough of an overview of the
program. The latter three elements can provide such an overview.
Pseudocode
Use of natural language, such as English, to describe algorithms often results in
a description that is too verbose and is subject to misinterpretation. To avoid
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dealing immediately with the possibly complicated syntax of the programming
language, we can instead use pseudocode, in which natural language and mathematical expressions are used to construct statements that look like computer
statements but without detailed syntax. Pseudocode may also use some simple
MATLAB syntax to explain the operation of the program.
As its name implies, pseudocode is an imitation of the actual computer code.
The pseudocode can provide the basis for comments within the program. In addition to providing documentation, pseudocode is useful for outlining a program
before writing the detailed code, which takes longer to write because it must conform to the strict rules of MATLAB.
Each pseudocode instruction may be numbered, but should be unambiguous
and computable. Note that MATLAB does not use line numbers except in the
Debugger. Each of the following examples illustrates how pseudocode can document each of the control structures used in algorithms: sequential, conditional,
and iterative operations.
Example 1. Sequential Operations Compute the perimeter p and the area A
of a triangle whose sides are a, b, and c. The formulas are
p
A = 1s(s - a)(s - b)(s - c)
p = a + b + c
s =
2
1. Enter the side lengths a, b, and c.
2. Compute the perimeter p.
p = a + b + c
3. Compute the semiperimeter s.
s =
4. Compute the area A.
p
2
A = 1s(s - a)(s - b)(s - c)
5. Display the results p and A.
6. Stop.
The program is
a = input(‘Enter the value of side a: ’);
b = input(‘Enter the value of side b: ’);
c = input(‘Enter the value of side c: ’);
p = a + b + c;
s = p/2;
A = sqrt(s*(s-a)*(s-b)*(s-c));
disp(‘The perimeter is:’)
p
disp(‘The area is:’)
A
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Program Design and Development
Example 2. Conditional Operations Given the (x, y) coordinates of a point,
compute its polar coordinates (r, ), where
y
r = 2x2 + y2 = tan - 1 a b
x
1. Enter the coordinates x and y.
2. Compute the hypoteneuse r.
r = sqrt(x^2+y^2)
3. Compute the angle .
3.1 If x 0
theta = atan(y/x)
3.2 Else
theta = atan(y/x) + pi
4. Convert the angle to degrees.
theta = theta*(180/pi)
5. Display the results r and theta.
6. Stop.
Note the use of the numbering scheme 3.1 and 3.2 to indicate subordinate
clauses. Note also that MATLAB syntax may be used for clarity where needed.
The following program implements the pseudocode using some of the MATLAB
features to be introduced in this chapter. It uses the relational operator >=, which
means “greater than or equal to” (in Section 4.2). The program also uses the
“if-else-end” construct, which is covered in Section 4.3.
x = input(‘Enter the value of x: ’);
y = input(‘Enter the value of y: ’);
r = sqrt(x^2+y^2);
if x >= 0
theta = atan(y/x);
else
theta = atan(y/x) + pi;
end
disp(‘The hypoteneuse is:’)
disp(r)
theta = theta*(180/pi);
disp(‘The angle is degrees is:’)
disp(theta)
Example 3. Iterative Operations Determine how many terms are required
for the sum of the series 10k2 - 4k + 2, k = 1, 2, 3, Á to exceed 20 000. What
is the sum for this many terms?
Because we do not know how many times we must evaluate the expression
10k2 - 4k + 2, we use a while loop, which is covered in Section 4.6.
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1. Initialize the total to zero.
2. Initialize the counter to zero.
3. While the total is less than 20 000 compute the total.
3.1 Increment the counter by 1.
k = k + 1
3.2 Update the total.
total = 10*k^2 - 4*k + 2 + total
4. Display the current value of the counter.
5. Display the value of the total.
6. Stop.
The following program implements the pseudocode. The statements in the
while loop are executed until the variable total equals or exceeds 2 104.
total = 0;
k = 0;
while total < 2e+4
k = k+1;
total = 10*k^2 - 4*k + 2 + total;
end
disp(‘The number of terms is:’)
disp(k)
disp(‘The sum is:’)
disp(total)
Finding Bugs
Debugging a program is the process of nding and removing the “bugs,” or errors,
in a program. Such errors usually fall into one of the following categories.
1. Syntax errors such as omitting a parenthesis or comma, or spelling a command name incorrectly. MATLAB usually detects the more obvious errors
and displays a message describing the error and its location.
2. Errors due to an incorrect mathematical procedure. These are called runtime
errors. They do not necessarily occur every time the program is executed;
their occurrence often depends on the particular input data. A common
example is division by zero.
The MATLAB error messages usually enable you to nd syntax errors. However ,
runtime errors are more dif cult to locate. To locate such an error, try the
following:
1. Always test your program with a simple version of the problem, whose
answers can be checked by hand calculations.
2. Display any intermediate calculations by removing semicolons at the end
of statements.
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Table 4.2–1 Relational operators
Relational operator
Meaning
<
<=
>
>=
==
~=
Less than.
Less than or equal to.
Greater than.
Greater than or equal to.
Equal to.
Not equal to.
3. To test user-de ned functions, try commenting out the function line and
running the le as a script.
4. Use the debugging features of the Editor/Debugger, which is discussed in
Section 4.8.
4.2 Relational Operators and Logical Variables
MATLAB has six relational operators to make comparisons between arrays.
These operators are shown in Table 4.2–1. Note that the equal to operator consists of two signs, not a single sign as you might expect. The single sign
is the assignment, or replacement, operator in MATLAB.
The result of a comparison using the relational operators is either 0 (if the
comparison is false) or 1 (if the comparison is true), and the result can be used as
a variable. For example, if x = 2 and y = 5, typing z = x < y returns the
value z = 1 and typing u = x == y returns the value u = 0. To make the
statements more readable, we can group the logical operations using parentheses.
For example, z = (x < y) and u = (x == y).
When used to compare arrays, the relational operators compare the arrays on
an element-by-element basis. The arrays being compared must have the same
dimension. The only exception occurs when we compare an array to a scalar. In
that case all the elements of the array are compared to the scalar. For example,
suppose that x = [6,3,9] and y = [14,2,9]. The following MATLAB
session shows some examples.
>>z = (x < y)
z =
1
0
0
>>z = (x ~= y)
z =
1
1
0
>>z = (x > 8)
z =
0
0
1
The relational operators can be used for array addressing. For example, with
x = [6,3,9] and y = [14,2,9], typing z = x(x < y) nds all the elements in x that are less than the corresponding elements in y. The result is z = 6.
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The arithmetic operators +, -, *, /, and \ have precedence over the relational
operators. Thus the statement z = 5 > 2 + 7 is equivalent to z = 5 >(2+7)
and returns the result z = 0. We can use parentheses to change the order of
precedence; for example, z = (5 > 2) + 7 evaluates to z = 8.
The relational operators have equal precedence among themselves, and
MATLAB evaluates them in order from left to right. Thus the statement
z = 5 > 3 ~= 1
is equivalent to
z = (5 > 3) ~= 1
Both statements return the result z = 0.
With relational operators that consist of more than one character, such as ==
or >=, be careful not to put a space between the characters.
The logical Class
When the relational operators are used, such as x = (5 > 2), they create
a logical variable, in this case, x. Prior to MATLAB 6.5, logical was an
attribute of any numeric data type. Now logical is a rst-class data type and
a MATLAB class, and so logical is now equivalent to other rst-class types
such as character and cell arrays. Logical variables may have only the values
1 (true) and 0 (false).
Just because an array contains only 0s and 1s, however, it is not necessarily a
logical array. For example, in the following session k and w appear the same, but
k is a logical array and w is a numeric array, and thus an error message is issued.
>>x = -2:2
x =
-2
-1
0
1
2
>>k = (abs(x)>1)
k =
1
0
0
0
1
>>z = x(k)
z =
-2
2
>>w = [1,0,0,0,1];
>>v = x(w)
??? Subscript indices must either be real positive...
integers or logicals.
The logical Function
Logical arrays can be created with the relational and logical operators and with
the logical function. The logical function returns an array that can be used
for logical indexing and logical tests. Typing B = logical(A), where A is a
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numeric array, returns the logical array B. So to correct the error in the previous
session, you may type instead w = logical([1,0,0,0,1]) before typing v = x(w).
When a nite, real value other than 1 or 0 is assigned to a logical variable, the
value is converted to logical 1 and a warning message is issued. For example, when
you type y = logical(9), y will be assigned the value logical 1 and a warning will be issued. You may use the double function to convert a logical array to
an array of class double. For example, x = (5 > 3); y = double(x);.
Some arithmetic operations convert a logical array to a double array. For example,
if we add zero to each element of B by typing B = B + 0, then B will be converted to a numeric (double) array. However, not all mathematical operations are dened for logical variables. For example, typing
>>x = ([2, 3] > [1, 6]);
>>y = sin(x)
will generate an error message. This is not an important issue because it hardly
makes sense to compute the sine of logical data or logical variables.
Accessing Arrays Using Logical Arrays
When a logical array is used to address another array, it extracts from that array
the elements in the locations where the logical array has 1s. So typing A(B),
where B is a logical array of the same size as A, returns the values of A at the
indices where B is 1.
Given A = [5,6,7;8,9,10;11,12,13] and B = logical
(eye(3)), we can extract the diagonal elements of A by typing C = A(B) to
obtain C = [5;9;13]. Specifying array subscripts with logical arrays extracts
the elements that correspond to the true (1) elements in the logical array.
Note, however, that using the numeric array eye(3), as C = A(eye(3)),
results in an error message because the elements of eye(3) do not correspond to
locations in A. If the numeric array values correspond to valid locations, you may
use a numeric array to extract the elements. For example, to extract the diagonal
elements of A with a numeric array, type C = A([1,5,9]).
MATLAB data types are preserved when indexed assignment is used. So
now that logical is a MATLAB data type, if A is a logical array, for example,
A = logical(eye(4)), then typing A(3,4) = 1 does not change A to a
double array. However, typing A(3,4) = 5 will set A(3,4) to logical 1 and
cause a warning to be issued.
4.3 Logical Operators and Functions
MATLAB has ve logical operators, which are sometimes called Boolean operators (see Table 4.3–1). These operators perform element-by-element operations.
With the exception of the NOT operator (~), they have a lower precedence than
the arithmetic and relational operators (see Table 4.3–2). The NOT symbol is
called the tilde.
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Table 4.3–1 Logical operators
Operator
Name
De nition
~
NOT
&
AND
|
OR
&&
Short-Circuit AND
||
Short-Circuit OR
~A returns an array of the same dimension as A; the new array has 1s where A is
0 and 0s where A is nonzero.
A & B returns an array of the same dimension as A and B; the new array has 1s
where both A and B have nonzero elements and 0s where either A or B is 0.
A | B returns an array of the same dimension as A and B; the new array has 1s
where at least one element in A or B is nonzero and 0s where A and B are both 0.
Operator for scalar logical expressions. A && B returns true if both A and B
evaluate to true, and false if they do not.
Operator for scalar logical expressions. A || B returns true if either A or B or
both evaluate to true, and false if they do not.
Table 4.3–2 Order of precedence for operator types
Precedence
Operator type
First
Second
Third
Fourth
Fifth
Parentheses; evaluated starting with the innermost pair.
Arithmetic operators and logical NOT (~); evaluated from left to right.
Relational operators; evaluated from left to right.
Logical AND.
Logical OR.
The NOT operation ~A returns an array of the same dimension as A; the new
array has 1s where A is 0 and 0s where A is nonzero. If A is logical, then ~A
replaces 1s with 0s and 0s with 1s. For example, if x = [0,3,9] and y =
[14,-2,9], then z = ~x returns the array z = [1,0,0] and the statement
u = ~x > y returns the result u = [0,1,0]. This expression is equivalent
to u = (~x) > y, whereas v = ~(x > y)gives the result v = [1,0,1].
This expression is equivalent to v = (x <= y).
The & and | operators compare two arrays of the same dimension. The only
exception, as with the relational operators, is that an array can be compared to a
scalar. The AND operation A & B returns 1s where both A and B have nonzero
elements and 0s where any element of A or B is 0. The expression z = 0 & 3
returns z = 0; z = 2 & 3 returns z = 1; z = 0 & 0 returns z = 0, and
z = [5,-3,0,0] & [2,4,0,5] returns z = [1,1,0,0]. Because of
operator precedence, z = 1 & 2 + 3 is equivalent to z = 1 & (2 + 3),
which returns z = 1. Similarly, z = 5 < 6 & 1 is equivalent to z =
(5 < 6) & 1, which returns z = 1.
Let x = [6,3,9] and y = [14,2,9] and let a = [4,3,12]. The
expression
z = (x > y) & a
gives z = [0,1,0], and
z = (x > y) & (x > a)
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159
returns the result z = [0,0,0]. This is equivalent to
z = x > y & x > a
which is much less readable.
Be careful when using the logical operators with inequalities. For example,
note that ~(x > y) is equivalent to x <= y. It is not equivalent to x < y. As
another example, the relation 5 < x < 10 must be written as
(5 < x) & (x < 10)
in MATLAB.
The OR operation A|B returns 1s where at least one of A and B has
nonzero elements and 0s where both A and B are 0. The expression z = 0|3
returns z = 1; the expression z = 0|0 returns z = 0; and
z = [5,-3,0,0]|[2,4,0,5]
returns z = [1,1,0,1]. Because of operator precedence,
z = 3 < 5|4 == 7
is equivalent to
z = (3 < 5)|(4 == 7)
which returns z = 1. Similarly, z = 1|0 & 1 is equivalent to z = (1|0) &
1, which returns z = 1, while z = 1|0 & 0 returns z = 0, and z = 0 &
0|1 returns z = 1.
Because of the precedence of the NOT operator, the statement
z = ~3 == 7|4 == 6
returns the result z = 0, which is equivalent to
z = ((~3) == 7)|(4 == 6)
The exclusive OR function xor(A,B) returns 0s where A and B are either
both nonzero or both 0, and 1s where either A or B is nonzero, but not both. The
function is de ned in terms of the AND, OR, and NOT operators as follows.
function z = xor(A,B)
z = (A|B) & ~(A & B);
The expression
z = xor([3,0,6],[5,0,0])
returns z = [0,0,1], whereas
z = [3,0,6]|[5,0,0]
returns z = [1,0,1].
Table 4.3–3 is a truth table that de nes the operations of the logical operators and the function xor. Until you acquire more experience with the logical
TRUTH TABLE
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Table 4.3–3 Truth table
x
y
~x
x|y
x & y
xor(x,y)
true
true
false
false
true
false
true
false
false
false
true
true
true
true
true
false
true
false
false
false
false
true
true
false
operators, you should use this table to check your statements. Remember that
true is equivalent to logical 1, and false is equivalent to logical 0. We can test the
truth table by building its numerical equivalent as follows. Let x and y represent
the rst two columns of the truth table in terms of 1s and 0s.
The following MATLAB session generates the truth table in terms of 1s
and 0s.
>>x = [1,1,0,0]’;
>>y = [1,0,1,0]’;
>>Truth_Table = [x,y,~x,x|y,x & y,xor(x,y)]
Truth_Table =
1 1 0 1 1 0
1 0 0 1 0 1
0 1 1 1 0 1
0 0 1 0 0 0
Starting with MATLAB 6, the AND operator (&) was given a higher precedence than the OR operator (|). This was not true in earlier versions of
MATLAB, so if you are using code created in an earlier version, you should
make the necessary changes before using it in MATLAB 6 or higher. For
example, now the statement y = 1|5 & 0 is evaluated as y = 1|(5 & 0),
yielding the result y = 1, whereas in MATLAB 5.3 and earlier, the statement
would have been evaluated as y = (1|5) & 0, yielding the result y = 0. To
avoid potential problems due to precedence, it is important to use parentheses in
statements containing arithmetic, relational, or logical operators, even where
parentheses are optional. MATLAB now provides a feature to enable the system
to produce either an error message or a warning for any expression containing &
and | that would be evaluated differently than in earlier versions. If you do not
use this feature, MATLAB will issue a warning as the default. To activate the
error feature, type feature(‘OrAndError’,1). To reinstate the default,
type feature(‘OrAndError’,0).
Short-Circuit Operators
The following operators perform AND and OR operations on logical expressions
containing scalar values only. They are called short-circuit operators because
they evaluate their second operand only when the result is not fully determined
by the rst operand. They are de ned as follows in terms of the two logical variables A and B.
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A && B Returns true (logical 1) if both A and B evaluate to true, and
false (logical 0) if they do not.
A||B Returns true (logical 1) if either A or B or both evaluate to true,
and false (logical 0) if they do not.
Thus in the statement A && B, if A equals logical zero, then the entire
expression will evaluate to false, regardless of the value of B, and therefore there
is no need to evaluate B.
For A||B, if A is true, regardless of the value of B, the statement will evaluate to true.
Table 4.3–4 lists several useful logical functions.
Table 4.3–4 Logical functions
Logical function
De nition
all(x)
Returns a scalar, which is 1 if all the elements in the vector x
are nonzero and 0 otherwise.
Returns a row vector having the same number of columns as
the matrix A and containing 1s and 0s, depending on whether
the corresponding column of A has all nonzero elements.
Returns a scalar, which is 1 if any of the elements in the
vector x is nonzero and 0 otherwise.
Returns a row vector having the same number of columns
as A and containing 1s and 0s, depending on whether the
corresponding column of the matrix A contains any nonzero
elements.
Computes an array containing the indices of the nonzero
elements of the array A.
Computes the arrays u and v containing the row and column
indices of the nonzero elements of the array A and computes
the array w containing the values of the nonzero elements.
The array w may be omitted.
Returns an array of the same dimension as A with 1s where
the elements of are nite and 0s elsewhere.
Returns a 1 if A is a character array and 0 otherwise.
Returns a 1 if A is an empty matrix and 0 otherwise.
Returns an array of the same dimension as A, with 1s where
A has ‘inf’ and 0s elsewhere.
Returns an array of the same dimension as A with 1s where
A has ‘NaN’ and 0s elsewhere. (‘NaN’ stands for “not a
number,” which means an unde ned result.)
Returns a 1 if A is a numeric array and 0 otherwise.
Returns a 1 if A has no elements with imaginary parts and
0 otherwise.
Converts the elements of the array A into logical values.
Returns an array the same dimension as A and B; the new
array has 1s where either A or B is nonzero, but not both,
and 0s where A and B are either both nonzero or both zero.
all(A)
any(x)
any(A)
nd(A)
[u,v,w] =
nd(A)
nite(A)
ischar(A)
isempty(A)
isinf(A)
isnan(A)
isnumeric(A)
isreal(A)
logical(A)
xor(A,B)
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Logical Operators and the nd Function
The nd function is very useful for creating decision-making programs, especially when combined with the relational or logical operators. The function
nd(x) computes an array containing the indices of the nonzero elements of the
array x. For example, consider the session
>>x = [-2, 0, 4];
>>y = nd(x)
y =
1
3
The resulting array y = [1, 3] indicates that the rst and third elements of x
are nonzero. Note that the nd function returns the indices, not the values. In the
following session, note the difference between the result obtained by
x(x < y) and the result obtained by nd (x < y).
>>x = [6, 3, 9, 11];y = [14, 2, 9, 13];
>>values = x(x < y)
values =
6
11
>>how_many = length (values)
how_many =
2
>>indices = nd(x < y)
indices =
1 4
Thus two values in the array x are less than the corresponding values in the array y.
They are the rst and fourth values, 6 and 1 1. To nd out how many , we could
also have typed length(indices).
The nd function is also useful when combined with the logical operators.
For example, consider the session
>>x = [5, -3, 0, 0, 8]; y = [2, 4, 0, 5, 7];
>>z = nd(x & y)
z =
1
2
5
The resulting array z = [1, 2, 5] indicates that the rst, second, and fth
elements of both x and y are nonzero. Note that the nd function returns the
indices, and not the values. In the following session, note the difference between
the result obtained by y(x & y) and the result obtained by nd(x & y) above.
>>x = [5, -3, 0, 0, 8];y = [2, 4, 0, 5, 7];
>>values = y(x & y)
values =
2
4
7
>>how_many = length(values)
how_many =
3
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163
Thus there are three nonzero values in the array y that correspond to nonzero values in the array x. They are the rst, second, and fth values, which are 2, 4, and 7.
In the above examples, there were only a few numbers in the arrays x and y,
and thus we could have obtained the answers by visual inspection. However,
these MATLAB methods are very useful either where there are so many data
that visual inspection would be very time-consuming, or where the values are
generated internally in a program.
Test Your Understanding
T4.3–1 If x = [5,-3,18,4] and y = [-9,13,7,4], what will be the
result of the following operations? Use MATLAB to check your answer.
a. z = ~y > x
b. z = x & y
c. z = x | y
d. z = xor(x,y)
T4.3–2 Suppose that x = [-9, -6, 0, 2, 5] and y = [-10, -6 2, 4,
6]. What is the result of the following operations? Determine the answers by hand, and then use MATLAB to check your answers.
a. z (x < y)
b. z (x > y)
c. z (x ~ y)
d. z (x y)
e. z (x > 2)
T4.3–3 Suppose that x = [-4, -1, 0, 2, 10] and y = [-5, -2, 2,
5, 9]. Use MATLAB to nd the values and the indices of the elements
in x that are greater than the corresponding elements in y.
Height and Speed of a Projectile
The height and speed of a projectile (such as a thrown ball) launched with a speed of 0
at an angle A to the horizontal are given by
h(t) = ␷0 t sin A - 0.5gt2
␷(t) = 2␷02 - 2␷0 gt sin A + g2t2
where g is the acceleration due to gravity. The projectile will strike the ground when
h(t) = 0, which gives the time to hit thit = 2(␷0>g)sin A. Suppose that A = 40°,
␷0 = 20 m/s, and g = 9.81 m/s2. Use the MATLAB relational and logical operators to
nd the times when the height is no less than 6 m and the speed is simultaneously no
greater than 16 m/s. In addition, discuss another approach to obtaining a solution.
EXAMPLE 4.3–1
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■ Solution
The key to solving this problem with relational and logical operators is to use the nd command to determine the times at which the logical expression (h >= 6) & (v <= 16)
is true. First we must generate the vectors h and v corresponding to times t1 and t2
between 0 … t … thit, using a spacing for time t that is small enough to achieve suf cient
accuracy for our purposes. We will choose a spacing of thit >100, which provides 101 values of time. The program follows. When computing the times t1 and t2, we must subtract
1 from u(1) and from length(u) because the rst element in the array t corresponds
to t = 0 (that is, t(1) is 0).
% Set the values for initial speed, gravity, and angle.
v0 = 20; g = 9.81; A = 40*pi/180;
% Compute the time to hit.
t_hit = 2*v0*sin(A)/g;
% Compute the arrays containing time, height, and speed.
t = 0:t_hit/100:t_hit;
h = v0*t*sin(A) - 0.5*g*t.^2;
v = sqrt(v0^2 - 2*v0*g*sin(A)*t + g^2*t.^2);
% Determine when the height is no less than 6
% and the speed is no greater than 16.
u = nd(h >= 6 & v <= 16);
% Compute the corresponding times.
t_1 = (u(1)- 1)*(t_hit/100)
t_2 = u(length(u)- 1)*(t_hit/100)
The results are t1 = 0.8649 and t2 = 1.7560. Between these two times h Ú 6 m and
␷ … 16 m/s.
We could have solved this problem by plotting h(t) and ␷(t), but the accuracy of the
results would be limited by our ability to pick points off the graph; in addition, if we had
to solve many such problems, the graphical method would be more time-consuming.
Test Your Understanding
T4.3–4 Consider the problem given in Example 4.3–1. Use relational and logical
operators to nd the times for which either the projectile’ s height is less
than 4 m or the speed is greater than 17 m/s. Plot h(t) and ␷(t) to con rm
your answer.
4.4 Conditional Statements
In everyday language we describe our decision making by using conditional
phrases such as “If I get a raise, I will buy a new car.” If the statement “I get a
raise” is true, the action indicated (buy a new car) will be executed. Here is another example: If I get at least a $100 per week raise, I will buy a new car; else,
I will put the raise into savings. A slightly more involved example is: If I get at
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least a $100 per week raise, I will buy a new car; else, if the raise is greater than
$50, I will buy a new stereo; otherwise, I will put the raise into savings.
We can illustrate the logic of the rst example as follows:
If I get a raise,
I will buy a new car
. (period)
Note how the period marks the end of the statement.
The second example can be illustrated as follows:
If I get at least a $100 per week raise,
I will buy a new car;
else,
I will put the raise into savings
. (period)
The third example follows.
If I get at least a $100 per week raise,
I will buy a new car;
else, if the raise is greater than $50,
I will buy a new stereo;
otherwise,
I will put the raise into savings
. (period)
The MATLAB conditional statements enable us to write programs that make
decisions. Conditional statements contain one or more of the if, else, and
elseif statements. The end statement denotes the end of a conditional statement, just as the period was used in the preceding examples. These conditional
statements have a form similar to the examples, and they read somewhat like
their English-language equivalents.
The if Statement
The if statement’s basic form is
if logical expression
statements
end
Every if statement must have an accompanying end statement. The end statement marks the end of the statements that are to be executed if the logical expression is true. A space is required between the if and the logical expression, which
may be a scalar, a vector, or a matrix.
For example, suppose that x is a scalar and that we want to compute y = 1x
only if x Ú 0. In English, we could specify this procedure as follows: If x is
greater than or equal to zero, compute y from y = 1x. The following if
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statement implements this procedure in MATLAB, assuming x already has a
scalar value.
if x >= 0
y = sqrt(x)
end
If x is negative, the program takes no action. The logical expression here is
x >= 0, and the statement is the single line y = sqrt(x).
The if structure may be written on a single line; for example,
if x >= 0, y = sqrt(x), end
However, this form is less readable than the previous form. The usual practice is
to indent the statements to clarify which statements belong to the if and its corresponding end and thereby improve readability.
The logical expression may be a compound expression; the statements may
be a single command or a series of commands separated by commas or semicolons or on separate lines. For example, if x and y have scalar values,
z = 0;w = 0;
if (x >= 0)&(y >= 0)
z = sqrt(x) + sqrt(y)
w = sqrt(x*y)
end
The values of z and w are computed only if both x and y are nonnegative. Otherwise, z and w retain their values of zero. The owchart is shown in Figure 4.4–1.
We may “nest” if statements, as shown by the following example.
if logical expression 1
statement group 1
if logical expression 2
statement group 2
end
end
Note that each if statement has an accompanying end statement.
The else Statement
When more than one action can occur as a result of a decision, we can use the
else and elseif statements along with the if statement. The basic structure
for the use of the else statement is
if logical expression
statement group 1
else
statement group 2
end
Figure 4.4–2 shows the owchart of this structure.
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Conditional Statements
4.4
Start
x ,y
False
x ≥ 0?
True
y ≥ 0?
167
Logical
Expression
False
True
False
Statement
Group 1
Statement
Group 2
True
Compute z , w
End
Figure 4.4–2 Flowchart of the else structure.
End
Figure 4.4–1 Flowchart illustrating
two logical tests.
For example, suppose that y = 1x for x Ú 0 and that y = ex - 1 for x 6 0.
The following statements will calculate y, assuming that x already has a scalar value.
if x >= 0
y = sqrt(x)
else
y = exp(x) - 1
end
When the test, if logical expression, is performed, where the logical
expression may be an array, the test returns a value of true only if all the elements
of the logical expression are true! For example, if we fail to recognize how the test
works, the following statements do not perform the way we might expect.
x = [4,-9,25];
if x < 0
disp(‘Some of the elements of x are negative.’)
else
y = sqrt(x)
end
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When this program is run, it gives the result
y =
2
0 + 3.000i
5
The program does not test each element in x in sequence. Instead it tests the truth
of the vector relation x < 0. The test if x < 0 returns a false value because
it generates the vector [0,1,0]. Compare the preceding program with the
following program.
x = [4,-9,25];
if x >= 0
y = sqrt(x)
else
disp(‘Some of the elements of x are negative.’)
end
When executed, it produces the following result: Some of the elements
of x are negative. The test if x < 0 is false, and the test if x >= 0
also returns a false value because x >= 0 returns the vector [1,0,1].
We sometimes must choose between a program that is concise, but perhaps
more dif cult to understand, and one that uses more statements than is necessary .
For example, the statements
if logical expression 1
if logical expression 2
statements
end
end
can be replaced with the more concise program
if logical expression 1 & logical expression 2
statements
end
The elseif Statement
The general form of the if statement is
if logical expression 1
statement group 1
elseif logical expression 2
statement group 2
else
statement group 3
end
The else and elseif statements may be omitted if not required. However, if
both are used, the else statement must come after the elseif statement to
take care of all conditions that might be unaccounted for. Figure 4.4–3 is the
owchart for the general if structure.
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4.4
Conditional Statements
Start
Logical
Expression 1
False
True
Statement
Group 1
Logical
Expression 2
False
True
Statement
Group 2
Statement
Group 3
End
Figure 4.4–3 Flowchart for the general if structure.
For example, suppose that y = ln x if x Ú 5 and that y = 1x if 0 … x 6 5.
The following statements will compute y if x has a scalar value.
if x >= 5
y = log(x)
else
if x >= 0
y = sqrt(x)
end
end
If x = - 2, for example, no action will be taken. If we use an elseif, we need
fewer statements. For example,
if x >= 5
y = log(x)
elseif x >= 0
y = sqrt(x)
end
Note that the elseif statement does not require a separate end statement.
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The else statement can be used with elseif to create detailed decisionmaking programs. For example, suppose that y = ln x for x 7 10 , y = 1x for
0 … x … 10, and y = ex - 1 for x 6 0. The following statements will compute
y if x already has a scalar value.
if x > 10
y = log(x)
elseif x >= 0
y = sqrt(x)
else
y = exp(x) - 1
end
Decision structures may be nested; that is, one structure can contain another
structure, which in turn can contain another, and so on. Indentations are used to
emphasize the statement groups associated with each end statement.
Test Your Understanding
T4.4–1 Given a number x and the quadrant q (q 1, 2, 3, 4), write a program
to compute sin1(x) in degrees, taking into account the quadrant. The
program should display an error message if x 7 1.
Checking the Number of Input and Output Arguments
Sometimes you will want to have a function act differently depending on how
many inputs it has. You can use the function nargin, which stands for “number
of input arguments.” Within the function you can use conditional statements to
direct the ow of the computation depending on how many input ar guments
there are. For example, suppose you want to compute the square root of the input
if there is only one, but compute the square root of the average if there are two
inputs. The following function does this.
function z = sqrtfun(x, y)
if (nargin == 1)
z = sqrt(x);
elseif (nargin == 2)
z = sqrt((x + y)/2);
end
The nargout function can be used to determine the number of output
arguments.
Strings and Conditional Statements
A string is a variable that contains characters. Strings are useful for creating input
prompts and messages and for storing and operating on data such as names and
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addresses. To create a string variable, enclose the characters in single quotes. For
example, the string variable name is created as follows:
>>name = ‘Leslie Student’
name =
Leslie Student
The following string, called number,
>>number = ‘123’
number =
123
is not the same as the variable number created by typing number = 123.
Strings are stored as row vectors in which each column represents a character. For example, the variable name has 1 row and 14 columns (each blank space
occupies one column). We can access any column the way we access any other
vector. For example, the letter S in the name Leslie Student occupies the eighth
column in the vector name. It can be accessed by typing name(8).
One of the most important applications for strings is to create input prompts
and output messages. The following prompt program uses the isempty(x)
function, which returns a 1 if the array x is empty and 0 otherwise. It also uses
the input function, whose syntax is
x = input(‘prompt’, ‘string’)
This function displays the string prompt on the screen, waits for input from the
keyboard, and returns the entered value in the string variable x. The function
returns an empty matrix if you press the Enter key without typing anything.
The following prompt program is a script le that allows the user to answer
Yes by typing either Y or y or by pressing the Enter key. Any other response is
treated as a No answer.
response = input(‘Do you want to continue? Y/N [Y]: ’,‘s’);
if (isempty(response))|(response == ‘Y’)|(response == ‘y’)
response = ‘Y’
else
response = ‘N’
end
Many more string functions are available in MATLAB. Type help strfun
to obtain information on these.
4.5 for Loops
A loop is a structure for repeating a calculation a number of times. Each repetition of the loop is a pass. MATLAB uses two types of explicit loops: the for
loop, when the number of passes is known ahead of time, and the while loop,
when the looping process must terminate when a speci ed condition is satis ed
and thus the number of passes is not known in advance.
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A simple example of a for loop is
for k = 5:10:35
x = k^2
end
The loop variable k is initially assigned the value 5, and x is calculated from
x = k^2. Each successive pass through the loop increments k by 10 and calculates x until k exceeds 35. Thus k takes on the values 5, 15, 25, and 35; and x
takes on the values 25, 225, 625, and 1225. The program then continues to execute any statements following the end statement.
The typical structure of a for loop is
for loop variable m:s:n
statements
end
Start
Set k = m
k > n?
Increment k
by s
False
Statements
End
Statements
following
the End
statement
Figure 4.5–1 Flowchart of a for loop.
True
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4.5
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173
The expression m:s:n assigns an initial value of m to the loop variable, which is
incremented by the value s, called the step value or incremental value. The statements are executed once during each pass, using the current value of the loop
variable. The looping continues until the loop variable exceeds the terminating
value n. For example, in the expression for k = 5:10:36, the nal value of k
is 35. Note that we need not place a semicolon after the for m:s:n statement
to suppress printing k. Figure 4.5–1 shows the owchart of a for loop.
Note that a for statement needs an accompanying end statement. The end
statement marks the end of the statements that are to be executed. A space is required between the for and the loop variable, which may be a scalar, a vector,
or a matrix, although the scalar case is by far the most common.
The for loop may be written on a single line; for example,
for x = 0:2:10, y = sqrt(x), end
However, this form is less readable than the previous form. The usual practice is
to indent the statements to clarify which statements belong to the for and its
corresponding end and thereby improve readability.
Series Calculation with a for Loop
EXAMPLE 4.5–1
Write a script le to compute the sum of the rst 15 terms in the series 5 k2 2k, k 1, 2, 3, . . . , 15.
■ Solution
Because we know how many times we must evaluate the expression 5k2 2k, we can use
a for loop. The script le is the following:
total = 0;
for k = 1:15
total = 5*k^2 - 2*k + total;
end
disp (‘The sum for 15 terms is:’)
disp (total)
The answer is 5960.
Plotting with a for Loop
EXAMPLE 4.5–2
Write a script le to plot the function
for 5
x
30.
1514x + 10
y = 10x + 10
L 10
x Ú 9
0 … x 6 9
x 6 0
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■ Solution
We choose a spacing dx 35/300 to obtain 301 points, which is suf cient to obtain a
smooth plot. The script le is the following:
dx = 35/300;
x = -5:dx:30;
for k = 1:length(x)
if x(k) >= 9
y(k) = 15*sqrt(4*x(k)) + 10;
elseif x(k) >= 0
y(k) = 10*x(k) + 10;
else
y(k) = 10;
end
end
plot (x,y), xlabel(’x’), ylabel(‘y’)
Note that we must use the index k to refer to x within the loop, as x(k).
NESTED LOOPS
We may nest loops and conditional statements, as shown by the following
example. (Note that each for and if statement needs an accompanying end
statement.)
Suppose we want to create a special square matrix that has 1s in the rst row
and rst column, and whose remaining elements are the sum of two elements, the
element above and the element to the left, if the sum is less than 20. Otherwise,
the element is the maximum of those two element values. The following function
creates this matrix. The row index is r; the column index is c. Note how indenting improves the readability.
function A = specmat(n)
A = ones(n);
for r = 1:n
for c = 1:n
if (r > 1) & (c > 1)
s = A(r-1,c) + A(r,c-1);
if s < 20
A(r,c) = s;
else
A(r,c) = max(A(r-1,c),A(r,c-1));
end
end
end
end
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Typing specmat(5) produces the following matrix:
1 1
1
1
1
1 2
3
4
5
£ 1 3 6 10 15 §
1 4 10 10 15
1 5 15 15 15
Test Your Understanding
T4.5–1 Write a script le using conditional statements to evaluate the following
function, assuming that the scalar variable x has a value. The function is
y = 1 x2 + 1 for x 0, y 3x 1 for 0 x 10, and y 9 sin (5x 50) 31 for x 10. Use your le to evaluate y for x 5,
x 5, and x 15, and check the results by hand.
T4.5–2 Use a for loop to determine the sum of the rst 20 terms in the series
3k2, k 1, 2, 3, . . . , 20. (Answer: 8610.)
T4.5–3 Write a program to produce the following matrix:
4
10
A = ≥
16
22
8
14
20
26
12
18
¥
24
30
Note the following rules when using for loops with the loop variable expression k = m:s:n:
■
■
■
■
■
■
The step value s may be negative. For example, k = 10:-2:4 produces
k 10, 8, 6, 4.
If s is omitted, the step value defaults to 1.
If s is positive, the loop will not be executed if m is greater than n.
If s is negative, the loop will not be executed if m is less than n.
If m equals n, the loop will be executed only once.
If the step value s is not an integer, round-off errors can cause the loop to
execute a different number of passes than intended.
When the loop is completed, k retains its last value. You should not alter the
value of the loop variable k within the statements. Doing so can cause unpredictable results.
A common practice in traditional programming languages such as BASIC
and FORTRAN is to use the symbols i and j as loop variables. However, this
convention is not good practice in MATLAB, which uses these symbols for the
imaginary unit 1 - 1.
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The break and continue Statements
It is permissible to use an if statement to “jump” out of the loop before the loop
variable reaches its terminating value. The break command, which terminates the
loop but does not stop the entire program, can be used for this purpose. For example,
for k = 1:10
x = 50 - k^2;
if x < 0
break
end
y = sqrt(x)
end
% The program execution jumps to here
% if the break command is executed.
However, it is usually possible to write the code to avoid using the break command. This can often be done with a while loop as explained in the next section.
The break statement stops the execution of the loop. There can be applications where we want to not execute the case producing an error but continue executing the loop for the remaining passes. We can use the continue statement
to do this. The continue statement passes control to the next iteration of the
for or while loop in which it appears, skipping any remaining statements in
the body of the loop. In nested loops, continue passes control to the next iteration of the for or while loop enclosing it.
For example, the following code uses a continue statement to avoid computing the logarithm of a negative number.
x = [10,1000,-10,100];
y = NaN*x;
for k = 1:length(x)
if x(k) < 0
continue
end
kvalue(k) = k;
y(k) = log10(x(k));
end
kvalue
y
The results are k = 1, 2, 0, 4 and y = 1, 3, NaN, 2.
Using an Array as a Loop Index
It is permissible to use a matrix expression to specify the number of passes. In
this case the loop variable is a vector that is set equal to the successive columns
of the matrix expression during each pass. For example,
A = [1,2,3;4,5,6];
for v = A
disp(v)
end
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is equivalent to
A = [1,2,3;4,5,6];
n = 3;
for k = 1:n
v = A(:,k)
end
The common expression k = m:s:n is a special case of a matrix expression in
which the columns of the expression are scalars, not vectors.
For example, suppose we want to compute the distance from the origin to a
set of three points speci ed by their xy coordinates (3, 7), (6,6), and (2,8). We
can arrange the coordinates in the array coord as follows.
c
3
7
6
6
2
d
8
Then coord = [3,6,2;7,6,8]. The following program computes the
distance and determines which point is farthest from the origin. The rst time
through the loop the index coord is [3, 7]’. The second time the index is
[6, 6]’, and during the nal pass it is [2, 8]’.
k = 0;
for coord = [3,6,2;7,6,8]
k = k + 1;
distance(k) = sqrt(coord’*coord)
end
[max_distance,farthest] = max(distance)
The previous program illustrates the use of an array index, but the problem
can be solved more concisely with the following program, which uses the diag
function to extract the diagonal elements of an array.
coord = [3,6,2;7,6,8];
distance = sqrt(diag(coord’*coord))
[max_distance,farthest] = max(distance)
Implied Loops
Many MATLAB commands contain implied loops. For example, consider these
statements.
x = [0:5:100];
y = cos(x);
To achieve the same result using a for loop, we must type
for k = 1:21
x = (k - 1)*5;
y(k) = cos(x);
end
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The nd command is another example of an implied loop. The statement y =
nd(x>0) is equivalent to
m = 0;
for k = 1:length(x)
if x(k) > 0
m = m + 1;
y(m) = k;
end
end
If you are familiar with a traditional programming language such as FORTRAN
or BASIC, you might be inclined to solve problems in MATLAB using loops, instead of using the powerful MATLAB commands such as nd. To use these
commands and to maximize the power of MATLAB, you might need to adopt a
new approach to problem solving. As the preceding example shows, you often
can save many lines of code by using MATLAB commands, instead of using
loops. Your programs will also run faster because MATLAB was designed for
high-speed vector computations.
Test Your Understanding
T4.5–4 Write a for loop that is equivalent to the command sum(A), where A
is a matrix.
Data Sorting
EXAMPLE 4.5–3
A vector x has been obtained from measurements. Suppose we want to consider any data
value in the range 0.1 x 0.1 as being erroneous. We want to remove all such elements and replace them with zeros at the end of the array. Develop two ways of doing
this. An example is given in the following table.
x(1)
x(2)
x(3)
x(4)
x(5)
x(6)
x(7)
■ Solution
Before
After
1.92
0.05
2.43
0.02
0.09
0.85
0.06
1.92
2.43
0.85
0
0
0
0
The following script le uses a for loop with conditional statements. Note how the
null array [] is used.
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x = [1.92,0.05,-2.43,-0.02,0.09,0.85,-0.06];
y = [];z = [];
for k = 1:length(x)
if abs(x(k)) >= 0.1
y = [y,x(k)];
else
z = [z,x(k)];
end
end
xnew = [y,zeros(size(z))]
The next script le uses the
nd function.
x = [1.92,0.05,-2.43,-0.02,0.09,0.85,-0.06];
y = x( nd(abs(x) >= 0.1));
z = zeros(size( nd(abs(x) < 0.1)));
xnew = [y,z]
Use of Logical Arrays as Masks
Consider the array A.
0
A =
9
J
- 34
-1
-14
49
4
25
K
64
The following program computes the array B by computing the square roots of
all the elements of A whose value is no less than 0 and adding 50 to each element
that is negative.
A = [0, -1, 4; 9, -14, 25; -34, 49, 64];
for m = 1:size(A,1)
for n = 1:size(A,2)
if A(m,n) >= 0
B(m,n) = sqrt(A(m,n));
else
B(m,n) = A(m,n) + 50;
end
end
end
B
The result is
B =
0
3
J
16
49
36
7
2
5
K
8
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CHAPTER 4 Programming with MATLAB
MASKS
When a logical array is used to address another array, it extracts from that
array the elements in the locations where the logical array has 1s. We can often
avoid the use of loops and branching and thus create simpler and faster programs
by using a logical array as a mask that selects elements of another array. Any
elements not selected will remain unchanged.
The following session creates the logical array C from the numeric array A
given previously.
>>A = [0, -1, 4; 9, -14, 25; -34, 49, 64];
>>C = (A >= 0);
The result is
1
C = 1
J
0
0
0
1
1
1
K
1
We can use this technique to compute the square root of only those elements
of A given in the previous program that are no less than 0 and add 50 to those
elements that are negative. The program is
A = [0, -1, 4; 9, -14, 25; -34, 49, 64];
C = (A >= 0);
A(C) = sqrt(A(C))
A(~C) = A(~C) + 50
The result after the third line is executed is
0
3
J
- 34
-1
-14
49
2
25
K
64
0
A =
3
J
16
49
36
7
2
5
K
8
A =
The result after the last line is executed is
EXAMPLE 4.5–4
Flight of an Instrumented Rocket
All rockets lose weight as they burn fuel; thus the mass of the system is variable. The following equations describe the speed y and height h of a rocket launched vertically,
neglecting air resistance. They can be derived from Newton’s law.
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y(t) = u ln
h(t) =
m0
- gt
m0 - qt
for Loops
(4.5–1)
u
(m - qt) ln(m0 - qt)
q 0
+ u(ln m0 + 1)t -
gt2
m0u
ln m0
2
q
(4.5–2)
where m0 is the rocket’s initial mass, q is the rate at which the rocket burns fuel mass, u is
the exhaust velocity of the burned fuel relative to the rocket, and g is the acceleration due
to gravity. Let b be the burn time, after which all the fuel is consumed. Thus the rocket’s
mass without fuel is me = m0 - qb.
For t 7 b the rocket engine no longer produces thrust, and the speed and height are
given by
␷(t) = ␷(b) - g(t - b)
(4.5–3)
2
h(t) = h(b) + ␷(b)(t - b) -
g(t - b)
2
(4.5–4)
The time tp to reach the peak height is found by setting y(t) = 0. The result is
tp = b + y(b)>g. Substituting this expression into the expression (4.5–4) for h(t) gives
the following expression for the peak height: hp = h(b) + y2(b)>(2g). The time at which
the rocket hits the ground is thit = tp + 12hp>g.
Suppose the rocket is carrying instruments to study the upper atmosphere, and we
need to determine the amount of time spent above 50 000 ft as a function of the burn time
b (and thus as a function of the fuel mass qb). Assume that we are given the following
values: me = 100 slugs, q = 1 slug/sec, u = 8000 ft/sec, and g = 32.2 ft/sec2. If the
rocket’s maximum fuel load is 100 slugs, the maximum value of b is 100>q = 100. Write
a MATLAB program to solve this problem.
Table 4.5–1 Pseudocode for Example 4.5–4
Enter data.
Increment burn time from 0 to 100. For each burn time value:
Compute m0, ␷b, hb, hp.
If hp ⱖ hdesired,
Compute tp, thit.
Increment time from 0 to thit.
Compute height as a function of time, using
the appropriate equation, depending on whether
burnout has occurred.
Compute the duration above desired height.
End of the time loop.
If hp ⬍ hdesired, set duration equal to zero.
End of the burn time loop.
Plot the results.
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■ Solution
Pseudocode for developing the program appears in Table 4.5–1. A for loop is a logical
choice to solve this problem because we know the burn time b and thit, the time it takes to
hit the ground. A MATLAB program to solve this problem appears in Table 4.5–2. It has
two nested for loops. The inner loop is over time and evaluates the equations of motion
at times spaced 1兾10 sec apart. This loop calculates the duration above 50 000 ft for a speci c value of the burn time b. We can obtain greater accuracy by using a smaller value of
the time increment dt. The outer loop varies the burn time in integer values from b = 1 to
b = 100. The nal result is the vector of durations for the various burn times. Figure 4.5–2
gives the resulting plot.
Table 4.5–2 MATLAB program for Example 4.5–4
% Script le rocket1.m
% Computes ight duration as a function of burn time.
% Basic data values.
m_e = 100; q = 1; u = 8000; g = 32.2;
dt = 0.1; h_desired = 50 000;
for b = 1:100 % Loop over burn time.
burn_time(b) = b;
% The following lines implement the formulas in the text.
m_0 = m_e + q*b; v_b = u*log(m_0/m_e) - g*b;
h_b = ((u*m_e)/q)*log(m_e/(m_e+q*b))+u*b - 0.5*g*b^2;
h_p = h_b + v_b^2/(2*g);
if h_p >= h_desired
% Calculate only if peak height > desired height.
t_p = b + v_b/g; % Compute peak time.
t_hit = t_p + sqrt(2*h_p/g); % Compute time to hit.
for p = 0:t_hit/dt
% Use a loop to compute the height vector.
k = p + 1; t = p*dt; time(k) = t;
if t <= b
% Burnout has not yet occurred.
h(k) = (u/q)*(m_0 - q*t)*log(m_0 - q*t)...
+ u*(log(m_0) + 1)*t - 0.5*g*t^2 ...
- (m_0*u/q)*log(m_0);
else
% Burnout has occurred.
h(k) = h_b - 0.5*g*(t - b)^2 + v_b*(t - b);
end
end
% Compute the duration.
duration(b) = length( nd(h>=h_desired))*dt;
else
% Rocket did not reach the desired height.
duration(b) = 0;
end
end % Plot the results.
plot(burn_time,duration),xlabel(‘Burn Time (sec)’),...
ylabel(‘Duration (sec)’),title(‘Duration Above 50 000 Feet’)
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Duration Above 50 000 ft
180
160
140
Duration (sec)
120
100
80
60
40
20
0
0
10
20
30
40
50
60
Burn Time (sec)
70
80
90
100
Figure 4.5–2 Duration above 50 000 ft as a function of the burn time.
4.6 while Loops
The while loop is used when the looping process terminates because a speci ed
condition is satis ed, and thus the number of passes is not known in advance.
A simple example of a while loop is
x = 5;
while x < 25
disp(x)
x = 2*x - 1;
end
The results displayed by the disp statement are 5, 9, and 17. The loop variable
x is initially assigned the value 5, and it has this value until the statement x =
2*x - 1 is encountered the rst time. The value then changes to 9. Before each
pass through the loop, x is checked to see whether its value is less than 25. If so,
the pass is made. If not, the loop is skipped and the program continues to execute
any statements following the end statement.
A principal application of while loops is when we want the loop to continue as long as a certain statement is true. Such a task is often more dif cult to
do with a for loop. The typical structure of a while loop follows.
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while logical expression
statements
end
MATLAB rst tests the truth of the logical expression. A loop variable must be included in the logical expression. For example, x is the loop variable in the statement while x < 25. If the logical expression is true, the statements are
executed. For the while loop to function properly, the following two conditions
must occur:
1. The loop variable must have a value before the while statement is
executed.
2. The loop variable must be changed somehow by the statements.
The statements are executed once during each pass, using the current value of
the loop variable. The looping continues until the logical expression is false.
Figure 4.6–1 shows the owchart of the while loop.
Start
Logical
Expression
False
True
Statements
(which increment
the loop variable)
End
Statements
following
the End
statement
Figure 4.6–1 Flowchart of the while loop.
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185
Each while statement must be matched by an accompanying end. As with
for loops, the statements should be indented to improve readability. You may
nest while loops, and you may nest them with for loops and if statements.
Always make sure that the loop variable has a value assigned to it before the
start of the loop. For example, the following loop can give unintended results if
x has an overlooked previous value.
while x < 10
x = x + 1;
y = 2*x;
end
If x has not been assigned a value prior to the loop, an error message will occur.
If we intend x to start at zero, then we should place the statement x = 0; before
the while statement.
It is possible to create an in nite loop, which is a loop that never ends. For
example,
x = 8;
while x ~= 0
x = x - 3;
end
Within the loop the variable x takes on the values 5, 2, 1, 4, . . . , and the condition x ~= 0 is always satis ed, so the loop never stops. If such a loop occurs,
press Ctrl-C to stop it.
Series Calculation with a while Loop
Write a script le to determine the number of terms required for the sum of the series
5k2 2k, k 1, 2, 3, . . . , to exceed 10 000. What is the sum for this many terms?
■ Solution
Because we do not know how many times we must evaluate the expression 5k2 2k, we
use a while loop. The script le is the following:
total = 0;
k = 0;
while total < 1e+4
k = k + 1;
total = 5*k^2 - 2*k + total;
end
disp(‘The number of terms is:’)
disp(k)
disp(‘The sum is:’)
disp(total)
The sum is 10 203 after 18 terms.
EXAMPLE 4.6–1
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Growth of a Bank Account
EXAMPLE 4.6–2
Determine how long it will take to accumulate at least $10 000 in a bank account if you
deposit $500 initially and $500 at the end of each year, if the account pays 5 percent annual interest.
■ Solution
Because we do not know how many years it will take, a while loop should be used. The
script le is the following.
amount = 500;
k=0;
while amount < 10000
k = k+1;
amount = amount*1.05 + 500;
end
amount
k
The nal results are amount 1.0789e 004, or $10 789, and k 14, or
14 years.
Time to Reach a Speci ed Height
EXAMPLE 4.6–3
Consider the variable-mass rocket treated in Example 4.5–4. Write a program to determine how long it takes for the rocket to reach 40 000 ft if the burn time is 50 sec.
■ Solution
The pseudocode appears in Table 4.6–1. Because we do not know the time required, a
while loop is convenient to use. The program in Table 4.6–2 performs the task and is
a modi cation of the program in Table 4.5–2. Note that the new program allows for the
possibility that the rocket might not reach 40 000 ft. It is important to write your programs to handle all such foreseeable circumstances. The answer given by the program
is 53 sec.
Table 4.6–1 Pseudocode for Example 4.6–3
Enter data.
Compute m0, ␷b, hb, hp.
If hp hdesired,
Use a while loop to increment time and compute height until desired height
is reached.
Compute height as a function of time, using the appropriate equation,
depending on whether burnout has occurred.
End of the time loop.
Display the results.
If hp hdesired, rocket cannot reach desired height.
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Table 4.6–2 MATLAB program for Example 4.6–3
% Script le rocket2.m
% Computes time to reach desired height.
% Set the data values.
h_desired = 40000; m_e = 100; q = 1;
u = 8000; g = 32.2; dt = 0.1; b = 50;
% Compute values at burnout, peak time, and height.
m_0 = m_e + q*b; v_b = u*log(m_0/m_e) - g*b;
h_b = ((u*m_e)/q)*log(m_e/(m_e+q*b))+u*b - 0.5*g*b^2;
t_p = b + v_b/g;
h_p = h_b + v_b^2/(2*g);
% If h_p > h_desired, compute time to reached h_desired.
if h_p > h_desired
h = 0; k = 0;
while h < h_desired % Compute h until h = h_desired.
t = k*dt; k = k + 1;
if t <= b
% Burnout has not yet occurred.
h = (u/q)*(m_0 - q*t)*log(m_0 - q*t)...
+ u*(log(m_0) + 1)*t - 0.5*g*t^2 ...
- (m_0*u/q)*log(m_0);
else
% Burnout has occurred.
h = h_b - 0.5*g*(t - b)^2 + v_b*(t - b);
end
end
% Display the results.
disp(‘The time to reach the desired height is:’)
disp(t)
else
disp(‘Rocket cannot achieve the desired height.’)
end
Test Your Understanding
T4.6–1 Use a while loop to determine how many terms in the series 3k2, k 1, 2, 3, . . . , are required for the sum of the terms to exceed 2000. What
is the sum for this number of terms? (Answer: 13 terms, with a sum
of 2457.)
T4.6–2 Rewrite the following code, using a while loop to avoid using the
break command.
for k = 1:10
x = 50 - k^2;
if x < 0
break
end
y = sqrt(x)
end
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T4.6–3 Find to two decimal places the largest value of x before the error in the
series approximation ex L 1 + x + x2>2 + x3>6 exceeds 1 percent.
(Answer: x = 0.83.)
4.7 The switch Structure
The switch structure provides an alternative to using the if, elseif, and
else commands. Anything programmed using switch can also be programmed using if structures. However, for some applications the switch
structure is more readable than code using the if structure. The syntax is
switch input expression (scalar or string)
case value1
statement group 1
case value2
statement group 2
.
.
.
otherwise
statement group n
end
The input expression is compared to each case value. If they are the same,
then the statements following that case statement are executed and processing continues with any statements after the end statement. If the input expression is a
string, then it is equal to the case value if strcmp returns a value of 1 (true). Only
the rst matching case is executed. If no match occurs, the statements following
the otherwise statement are executed. However, the otherwise statement is
optional. If it is absent, execution continues with the statements following the end
statement if no match exists. Each case value statement must be on a single line.
For example, suppose the variable angle has an integer value that represents an angle measured in degrees from North. The following switch block
displays the point on the compass that corresponds to that angle.
switch angle
case 45
disp(‘Northeast’)
case 135
disp(‘Southeast’)
case 225
disp(‘Southwest’)
case 315
disp(‘Northwest’)
otherwise
disp(‘Direction Unknown’)
end
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The switch Structure
The use of a string variable for the input expression can result in very
readable programs. For example, in the following code the numeric vector x
has values, and the user enters the value of the string variable response; its
intended values are min, max, or sum. The code then either finds the minimum or maximum value of x or sums the elements of x, as directed by the
user.
t = [0:100]; x = exp(-t).*sin(t);
response = input(‘Type min, max, or sum.’,’s’)
response = lower(‘response’);
switch response
case min
minimum = min(x)
case max
maximum = max(x)
case sum
total = sum(x)
otherwise
disp(‘You have not entered a proper choice.’)
end
The switch statement can handle multiple conditions in a single case
statement by enclosing the case value in a cell array. For example, the following
switch block displays the corresponding point on the compass, given the integer angle measured from North.
switch angle
case {0,360}
disp(‘North’)
case {-180,180}
disp(‘South’)
case {-270,90}
disp(‘East’)
case {-90,270}
disp(‘West’)
otherwise
disp(‘Direction Unknown’)
end
Test Your Understanding
T4.7–1 Write a program using the switch structure to input one angle, whose
value may be 45, 45, 135, or 135 , and display the quadrant (1, 2, 3,
or 4) containing the angle.
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Using the switch Structure for Calendar Calculations
EXAMPLE 4.7–1
Use the switch structure to compute the total elapsed days in a year, given the number
(1–12) of the month, the day, and an indication of whether the year is a leap year.
■ Solution
Note that February has an extra day if the year is a leap year. The following function
computes the total elapsed number of days in a year, given the month, the day of the
month, and the value of extra_day, which is 1 for a leap year and 0 otherwise.
function total_days = total(month,day,extra_day)
total_days = day;
for k = 1:month - 1
switch k
case {1,3,5,7,8,10,12}
total_days = total_days + 31;
case {4,6,9,11}
total_days = total_days + 30;
case 2
total_days = total_days + 28 + extra_day;
end
end
The function can be used as shown in the following program.
month = input(‘Enter month (1 - 12): ‘);
day = input(‘Enter day (1 - 31): ‘);
extra_day = input(‘Enter 1 for leap year; 0 otherwise: ‘);
total_days = total(month,day,extra_day)
One of the chapter problems for Section 4.4 (Problem 19) asks you to write
a program to determine whether a given year is a leap year.
4.8 Debugging MATLAB Programs
Use of the MATLAB Editor/Debugger as an M- le editor was discussed in Section 1.4. Figure 1.4–1 shows the Editor/Debugger screen. Figure 4.8–1 shows the
Debugger containing two programs to be analyzed. Here we discuss its use as a
debugger. Before you use the Debugger, try to debug your program using the
commonsense guidelines presented under Debugging Script Files in Section 1.4.
MATLAB programs are often short because of the power of the commands, and
you may not need to use the Debugger unless you are writing large programs.
However, the cell mode discussed in this section is useful even for short programs.
The Editor/Debugger menu bar contains the following items: File, Edit, Text,
Go, Cell, Tools, Debug, Desktop, Window, and Help. The File, Edit, Desktop,
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Figure 4.8–1 The Editor/Debugger containing two programs to be analyzed.
Window, and Help menus are similar to those in the Desktop. The Go menu lets
you go backward or forward in the le. The Cell menu will be discussed shortly.
The Tools menu involves advanced topics that will not be treated in this text. The
Desktop menu is similar to that in the Command window. It enables you to dock
and undock windows, arrange the Editor window, and turn the Editor toolbar on
and off.
Below the menu bar is the Editor/Debugger toolbar. It enables you to access
several of the items in the menus with one click of the mouse. Hold the mouse
cursor over a button on the toolbar to see its function. For example, clicking the
button with the binoculars icon is equivalent to selecting Find and Replace from
the Edit menu. One item on the toolbar that is not in the menus is the function
button with the script f icon ( f ). Use this button to go to a particular function in
the M- le. The list of functions that you will see includes only those functions
whose function statements are in the program. The list does not include functions
that are called from the M- le.
The Text menu supplements the Edit menu for creating M- les. With the
Text menu you can insert or remove comments, increase or decrease the amount
of indenting, turn on smart indenting, and evaluate and display the values of
selected variables in the Command window. Click anywhere in a previously
typed line, and then click Comment in the Text menu. This makes the entire line
a comment. To turn a commented line into an executable line, click anywhere in
the line, and then click Uncomment in the Text menu.
Cell Mode
The cell mode can be used to debug programs. It can also be used to generate
a report. See the end of Section 5.2 for a discussion of the latter usage. A cell is
a group of commands. (Such a cell should not be confused with the cell array
data type covered in Section 2.6.) Type the double percent character (%%) to
mark the beginning of a new cell; it is called a cell divider. The cell mode is enabled when you rst enter your program into the Editor . To disable cell mode,
click the Cell button and select Disable Cell Mode. The cell toolbar is shown
in Figure 4.8–2.
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Figure 4.8–2 The cell mode of the Editor/Debugger.
Consider the following simple program that plots either a quadratic or a
cubic function.
%% Evaluate a quadratic and a cubic.
clear, clc
x = linspace(0, 10, 300);
%% Quadratic
y1 = polyval([1, -8, 6], x); plot(x,y1)
%% Cubic
y2 = polyval([1, -11, 9, 9], x); plot(x,y2)
After entering and saving the program, you can click on one of the evaluation
icons shown on the left-hand side of the cell toolbar (see Figure 4.8–2). These
enable you to evaluate the current single cell (where the cursor is currently), to evaluate the current cell and advance to the next cell, or to evaluate the entire program.
A useful feature of cell mode is that it enables you to evaluate the results of
changing a parameter. For example, in Figure 4.8–2, suppose the cursor is next
to the number 8. If you click the plus ( ) or minus () sign in the cell toolbar,
the parameter (8) will be decremented or incremented by the increment shown
in the window (1.0 is the default, which you can change). If you have already run
the program and the quadratic plot is on the screen, click the minus sign once to
change the parameter from 8 to 9 and watch the plot change.
You can also change the parameter by a divisive or multiplicative factor (1.1
is the default). Click the divide or multiply symbol on the cell toolbar.
The Debug Menu
BREAKPOINT
Breakpoints are points in the file where execution stops temporarily so that
you can examine the values of the variables up to that point. You set breakpoints with the Set/Clear Breakpoint item on the Debug menu. Use the Step,
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193
Step In, and Step Out items on the Debug menu to step through your file
after you have set breakpoints and run the file. Click Step to watch the script
execute one step at a time. Click Step In to step into the first executable line
in a function being called. Click Step Out in a called function to run the rest
of the function and then return to the calling program.
The solid green arrow to the left of the line text indicates the next line to be
executed. When this arrow changes to a hollow green arrow, MATLAB control
is now in a function being called. Execution returns to the line with the solid
green arrow after the function completes its operation. The arrow turns yellow
at a line where execution pauses or where a function completes its operation.
When the program pauses, you can assign new values to a variable, using either the Command window or the Array Editor.
Click on the Go Until Cursor item to run the le until it reaches the line
where the cursor is; this process sets a temporary breakpoint at the cursor. You
can save and execute your program directly from the Debug menu if you want,
by clicking on Run (or Save and Run if you have made changes). You can also
click on the Run icon. You need not set any breakpoints beforehand. Click Exit
Debug Mode to return to normal editing. To save any changes you have made to
the program, rst exit the debug mode, and then save the le.
Using Breakpoints
Most debugging sessions start by setting a breakpoint. A breakpoint stops M- le
execution at a speci ed line and allows you to view or change values in the function’s workspace before resuming execution. To set a breakpoint, position the
cursor in the line of text and click on the breakpoint icon in the toolbar or select
Set/Clear Breakpoints from the Debug menu. You can also set a breakpoint by
right-clicking on the line of text to bring up the Context menu and choosing
Set/Clear Breakpoint. A red circle next to a line indicates that a breakpoint is set
at that line. If the line selected for a breakpoint is not an executable line, then the
breakpoint is set at the next executable line. The Debug menu enables you to
clear all the breakpoints (select Clear Breakpoints in All Files). The Debug
menu also lets you halt M- le execution if your code generates a warning, an
error, or an NaN or Inf value (select Stop if Errors/ Warnings).
4.9 Applications to Simulation
Simulation is the process of building and analyzing the output of computer programs that describe the operations of an organization, process, or physical system.
Such a program is called a computer model. Simulation is often used in operations
research, which is the quantitative study of an organization in action, to nd ways
to improve the functioning of the organization. Simulation enables engineers to
study the past, present, and future actions of the organization for this purpose.
Operations research techniques are useful in all engineering elds. Common
examples include airline scheduling, traf c ow studies, and production lines. The
MATLAB logical operators and loops are excellent tools for building simulation
programs.
OPERATIONS
RESEARCH
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A College Enrollment Model: Part I
EXAMPLE 4.9–1
As an example of how simulation can be used for operations research, consider the following college enrollment model. A certain college wants to analyze the effect of
admissions and freshman retention rate on the college’s enrollment so that it can predict
the future need for instructors and other resources. Assume that the college has estimates of the percentages of students repeating a grade or leaving school before graduating. Develop a matrix equation on which to base a simulation model that can help in
this analysis.
■ Solution
Suppose that the current freshman enrollment is 500 students and the college decides to
admit 1000 freshmen per year from now on. The college estimates that 10 percent of
the freshman class will repeat the year. The number of freshmen in the following year will
be 0.1(500) 1000 1050, then it will be 0.1(1050) 1000 1105, and so on. Let
x1(k) be the number of freshmen in year k, where k 1, 2, 3, 4, 5, 6, . . . . Then in year
k 1, the number of freshmen is given by
x1(k + 1) = 10 percent of previous freshman class
repeating freshman year
+ 1000 new freshmen
= 0.1x1(k) + 1000
(4.9–1)
Because we know the number of freshmen in the rst year of our analysis (which is
500), we can solve this equation step by step to predict the number of freshmen in the
future.
Let x2(k) be the number of sophomores in year k. Suppose that 15 percent of the
freshmen do not return and that 10 percent repeat freshman year. Thus 75 percent of
the freshman class returns as sophomores. Suppose also 5 percent of the sophomores
repeat the sophomore year and that 200 sophomores each year transfer from other
schools. Then in year k 1, the number of sophomores is given by
x2(k + 1) = 0.75x1(k) + 0.05x2(k) + 200
To solve this equation, we need to solve the “freshman” equation (4.9–1) at the same
time, which is easy to do with MATLAB. Before we solve these equations, let us develop
the rest of the model.
Let x3(k) and x4(k) be the number of juniors and seniors, respectively, in year k.
Suppose that 5 percent of the sophomores and juniors leave school and that 5 percent of
the sophomores, juniors, and seniors repeat the grade. Thus 90 percent of the sophomores
and juniors return and advance in grade. The models for the juniors and seniors are
x3(k + 1) = 0.9x2(k) + 0.05x3(k)
x4(k + 1) = 0.9x3(k) + 0.05x4(k)
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195
These four equations can be written in the following matrix form:
x1(k + 1)
0.1
0.75
x2(k + 1)
≥
¥≥
x3(k + 1)
0
x4(k + 1)
0
0
0.05
0.9
0
0
0
0.05
0.9
x1(k)
0
0
x (k)
¥ ≥ 2 ¥
x3(k)
0
0.05
x4(k)
1000
200
¥
≥
0
0
In Example 4.9–2 we will see how to use MATLAB to solve such equations.
Test Your Understanding
T4.9–1 Suppose that 70 percent of the freshmen, instead of 75 percent, return for
the sophomore year. How does the previous equation change?
A College Enrollment Model: Part II
EXAMPLE 4.9–2
To study the effects of admissions and transfer policies, generalize the enrollment model
in Example 4.9–1 to allow for varying admissions and transfers.
■ Solution
Let a(k) be the number of new freshmen admitted in the spring of year k for the following year k + 1, and let d(k) be the number of transfers into the following year’s sophomore
class. Then the model becomes
x1(k + 1) = c11x1(k) + a(k)
x2(k + 1) = c21x1(k) + c22x2(k) + d(k)
x3(k + 1) = c32x2(k) + c33x3(k)
x4(k + 1) = c43x3(k) + c44x4(k)
where we have written the coef cients c21, c22, and so on in symbolic, rather than numerical, form so that we can change their values if desired.
This model can be represented graphically by a state transition diagram, like the one
shown in Figure 4.9–1. Such diagrams are widely used to represent time-dependent and
probabilistic processes. The arrows indicate how the model’s calculations are updated for
Sophomore Transfers
d (k )
c 32
c 21
a (k )
c 43
New Admissions
x 1( k )
Freshmen
Juniors
Sophomores
c 11
x 3( k )
x 2( k )
c 22
x 4( k )
Seniors
c 33
Figure 4.9–1 The state transition diagram for the college enrollment model.
c 44
STATE
TRANSITION
DIAGRAM
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each new year. The enrollment at year k is described completely by the values of x1(k),
x2(k), x3(k), and x4(k), that is, by the vector x(k), which is called the state vector. The elements of the state vector are the state variables. The state transition diagram shows how
the new values of the state variables depend on both the previous values and the inputs
a(k) and d(k).
The four equations can be written in the following matrix form:
x1(k + 1)
c11
x (k + 1)
c
≥ 2
¥ ≥ 21
x3(k + 1)
0
x4(k + 1)
0
0
c22
c32
0
0
0
c33
c43
0
x1(k)
0
x (k)
¥ ≥ 2 ¥
x3(k)
0
x4(k)
c44
a(k)
d(k)
≥
¥
0
0
or more compactly as
x(k + 1) = Cx(k) + b(k)
where
x1(k)
x2(k)
x(k) = ≥
¥
x3(k)
x4(k)
a(k)
d(k)
b(k) = ≥
¥
0
0
and
c11
c21
C = ≥
0
0
0
c22
c32
0
0
0
c33
c43
0
0
¥
0
c44
Suppose that the initial total enrollment of 1480 consists of 500 freshmen, 400 sophomores, 300 juniors, and 280 seniors. The college wants to study, over a 10-year period,
the effects of increasing admissions by 100 each year and transfers by 50 each year until
the total enrollment reaches 4000; then admissions and transfers will be held constant.
Thus the admissions and transfers for the next 10 years are given by
a(k) = 900 + 100k
d(k) = 150 + 50k
for k 1, 2, 3, . . . until the college’s total enrollment reaches 4000; then admissions and
transfers are held constant at the previous year’s levels. We cannot determine when this
event will occur without doing a simulation. Table 4.9–1 gives the pseudocode for solving this problem. The enrollment matrix E is a 4 10 matrix whose columns represent
the enrollment in each year.
Because we know the length of the study (10 years), a for loop is a natural choice.
We use an if statement to determine when to switch from the increasing admissions
and transfer schedule to the constant schedule. A MATLAB script le to predict the enrollment for the next 10 years appears in Table 4.9–2. Figure 4.9–2 shows the resulting
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Applications to Simulation
Table 4.9–1 Pseudocode for Example 4.9–2
Enter the coef cient matrix C and the initial enrollment vector x.
Enter the initial admissions and transfers, a(1) and d(1).
Set the rst column of the enrollment matrix E equal to x.
Loop over years 2 to 10.
If the total enrollment is 4000, increase admissions by 100 and transfers
by 50 each year.
If the total enrollment is 4000, hold admissions and transfers constant.
Update the vector x, using x ⴝ Cx ⴙ b.
Update the enrollment matrix E by adding another column composed of x.
End of the loop over years 2 to 10.
Plot the results.
Table 4.9–2 College enrollment model
% Script le enroll1.m. Computes college enrollment.
% Model’s coef cients.
C = [0.1,0,0,0;0.75,0.05,0,0;0,0.9,0.05,0;0,0,0.9,0.05];
% Initial enrollment vector.
x = [500;400;300;280];
% Initial admissions and transfers.
a(1) = 1000; d(1) = 200;
% E is the 4 x 10 enrollment matrix.
E(:,1) = x;
% Loop over years 2 to 10.
for k = 2:10
% The following describes the admissions
% and transfer policies.
if sum(x) <= 4000
% Increase admissions and transfers.
a(k) = 900+100*k;
d(k) = 150+50*k;
else
% Hold admissions and transfers constant.
a(k) = a(k-1);
d(k) = d(k-1);
end
% Update enrollment matrix.
b = [a(k);d(k);0;0];
x = C*x+b;
E(:,k) = x;
end
% Plot the results.
plot(E’),hold,plot(E(1,:),’o’),plot(E(2,:),’+’),plot(E(3,:),’*’),...
plot(E(4,:),’x’),xlabel(‘Year’),ylabel(‘Number of Students’),...
gtext(‘Frosh’),gtext(‘Soph’),gtext(‘Jr’),gtext(‘Sr’),...
title(‘Enrollment as a Function of Time’)
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Enrollment As a Function of Time
1600
1400
Frosh
1200
Number of Students
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1000
Soph
800
Jr
Sr
600
400
200
1
2
3
4
5
Year
6
7
8
9
10
Figure 4.9–2 Class enrollments versus time.
plot. Note that after year 4 there are more sophomores than freshmen. The reason is
that the increasing transfer rate eventually overcomes the effect of the increasing
admission rate.
In actual practice this program would be run many times to analyze the effects of
different admissions and transfer policies and to examine what happens if different
values are used for the coef cients in the matrix C (indicating different dropout and
repeat rates).
Test Your Understanding
T4.9–2 In the program in Table 4.9–2, lines 16 and 17 compute the values of
a(k) and d(k). These lines are repeated here:
a(k) = 900 + 100 * k
d(k) = 150 + 50 * k;
Why does the program contain the line a(1)=1000; d(1)=200;?
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4.10
Summary
199
Table 4.10–1 Guide to MATLAB commands introduced in Chapter 4
Relational operators
Logical operators
Order of precedence for operator types
Truth table
Logical functions
Table 4.2–1
Table 4.3–1
Table 4.3–2
Table 4.3–3
Table 4.3–4
Miscellaneous commands
Command
Description
Section
break
case
continue
double
else
elseif
end
for
if
input(‘s1’, ‘s’)
logical
nargin
nargout
switch
Terminates the execution of a for or a while loop.
Used with switch to direct program execution.
Passes control to the next iteration of a for or while loop.
Converts a logical array to class double.
Delineates an alternate block of statements.
Conditionally executes statements.
Terminates for, while, and if statements.
Repeats statements a speci c number of times.
Executes statements conditionally.
Display the prompt string s1 and stores user input as a string.
Converts numeric values to logical values.
Determines the number of input arguments of a function.
Determines the number of output arguments of a function.
Directs program execution by comparing the input expression with
the associated case expressions.
Repeats statements an inde nite number of times.
Exclusive OR function.
4.5, 4.6
4.7
4.5, 4.6
4.2
4.4
4.4
4.4, 4.5, 4.6
4.5
4.4
4.4
4.2
4.4
4.4
while
xor
4.10 Summary
Now that you have nished this chapter, you should be able to write programs that
can perform decision-making procedures; that is, the program’s operations depend
on results of the program’s calculations or on input from the user. Sections 4.2,
4.3, and 4.4 covered the necessary functions: the relational operators, the logical
operators and functions, and the conditional statements.
You should also be able to use MATLAB loop structures to write programs
that repeat calculations a speci ed number of times or until some condition is
satis ed. This feature enables engineers to solve problems of great complexity or
requiring numerous calculations. The for loop and while loop structures were
covered in Sections 4.5 and 4.6. Section 4.7 covered the switch structure.
Section 4.8 gave an overview and an example of how to debug programs
using the Editor/Debugger. Section 4.9 presented an application of these methods
to simulation, which enables engineers to study the operation of complicated
systems, processes, and organizations.
Tables summarizing the MATLAB commands introduced in this chapter are
located throughout the chapter. Table 4.10–1 will help you locate these tables. It
also summarizes those commands not found in the other tables.
4.7
4.6
4.3
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Key Terms with Page References
Relational operator, 155
Simulation, 193
State transition diagram, 195
Structure chart, 150
Structured programming, 148
switch structure, 188
Top-down design, 149
Truth table, 159
while loop, 183
Breakpoint, 192
Conditional statement, 164
Flowchart, 150
for loop, 171
Implied loop, 177
Logical operator, 157
Masks, 180
Nested loops, 174
Operations research, 193
Pseudocode, 151
Problems
You can nd answers to problems marked with an asterisk at the end of the text.
Section 4.1
1.
The volume V and surface area A of a sphere of radius r are given by
4
V = r3
A = 4r 2
3
a. Develop a pseudocode description of a program to compute V and A
for 0 … r … 3 m and to plot V versus A.
b. Write and run the program described in part a.
2.
The roots of the quadratic equation ax2 + bx + c = 0 are given by
- b 1b2 - 4ac
2a
a. Develop a pseudocode description of a program to compute both roots
given the values of a, b, and c. Be sure to identify the real and
imaginary parts.
b. Write the program described in part a and test it for the following
cases:
1. a = 2, b = 10, c = 12
2. a = 3, b = 24, c = 48
3. a = 4, b = 24, c = 100
x =
3.
It is desired to compute the sum of the rst 10 terms of the series
14k3 - 20k2 + 5k, k = 1, 2, 3, Á
a. Develop a pseudocode description of the required program.
b. Write and run the program described in part a.
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Problems
Section 4.2
4.*
Suppose that x = 6. Find the results of the following operations by hand
and use MATLAB to check your results.
a. z = (x < 10)
b. z = (x == 10)
c. z = (x >= 4)
d. z = (x ~= 7)
5.* Find the results of the following operations by hand and use MATLAB to
check your results.
a. z = 6 > 3 + 8
b. z = 6 + 3 > 8
c. z = 4 > (2 + 9)
d. z = (4 < 7) + 3
e. z = 4 < 7 + 3
f. z = (4<7)*5
g. z = 4<(7*5)
h. z = 2/5>=5
6.* Suppose that x = [10, -2, 6, 5, -3] and y = [9, -3, 2,
5, -1]. Find the results of the following operations by hand and use
MATLAB to check your results.
a. z = (x < 6)
b. z = (x <= y)
c. z = (x == y)
d. z = (x ~= y)
7. For the arrays x and y given below, use MATLAB to nd all the elements
in x that are greater than the corresponding elements in y.
x = [-3, 0, 0, 2, 6, 8]
y = [-5, -2, 0, 3, 4, 10]
8. The array price given below contains the price in dollars of a certain
stock over 10 days. Use MATLAB to determine how many days the price
was above $20.
price = [19, 18, 22, 21, 25, 19, 17, 21, 27, 29]
9. The arrays price_A and price_B given below contain the price in
dollars of two stocks over 10 days. Use MATLAB to determine how
many days the price of stock A was above the price of stock B.
price_A = [19, 18, 22, 21, 25, 19, 17, 21, 27, 29]
10.
price_B = [22, 17, 20, 19, 24, 18, 16, 25, 28, 27]
The arrays price_A, price_B, and price_C given below contain the
price in dollars of three stocks over 10 days.
a. Use MATLAB to determine how many days the price of stock A was
above both the price of stock B and the price of stock C.
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b. Use MATLAB to determine how many days the price of stock A was
above either the price of stock B or the price of stock C.
c. Use MATLAB to determine how many days the price of stock A was
above either the price of stock B or the price of stock C, but not
both.
price_A = [19, 18, 22, 21, 25, 19, 17, 21, 27, 29]
price_B = [22, 17, 20, 19, 24, 18, 16, 25, 28, 27]
price_C = [17, 13, 22, 23, 19, 17, 20, 21, 24, 28]
Section 4.3
11.* Suppose that x = [-3, 0, 0, 2, 5, 8] and y = [-5, -2,
0, 3, 4, 10]. Find the results of the following operations by hand
and use MATLAB to check your results.
a. z = y < ~ x
b. z = x & y
c. z = x|y
d. z = xor(x,y)
12. The height and speed of a projectile (such as a thrown ball) launched with
a speed of y0 at an angle A to the horizontal are given by
h(t) = ␷0t sin A - 0.5 gt2
␷(t) = 1 ␷20 - 2␷0gt sin A + g2t2
where g is the acceleration due to gravity. The projectile will strike the
ground when h(t) = 0, which gives the time to hit thit = 2(␷0>g) sin A.
Suppose that A 30°, y0 = 40 m/s, and g = 9.81m/s2. Use the
MATLAB relational and logical operators to nd the times when
a. The height is no less than 15 m.
b. The height is no less than 15 m and the speed is simultaneously no
greater than 36 m/s.
c. The height is less than 5 m or the speed is greater than 35 m/s.
13.* The price, in dollars, of a certain stock over a 10-day period is given in
the following array.
price = [19, 18, 22, 21, 25, 19, 17, 21, 27, 29]
Suppose you owned 1000 shares at the start of the 10-day period, and you
bought 100 shares every day the price was below $20 and sold 100 shares
every day the price was above $25. Use MATLAB to compute (a) the
amount you spent in buying shares, (b) the amount you received from the
sale of shares, (c) the total number of shares you own after the 10th day,
and (d ) the net increase in the worth of your portfolio.
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Problems
14.
Let e1 and e2 be logical expressions. DeMorgan’s laws for logical
expressions state that
NOT(e1 AND e2) implies that (NOT e1) OR (NOT e2)
and
NOT(e1 OR e2) implies that (NOT e1) AND (NOT e2)
Use these laws to nd an equivalent expression for each of the following
expressions and use MATLAB to verify the equivalence.
15.
a. ~((x < 10) & (x >= 6))
b. ~((x == 2) | (x > 5))
Are these following expressions equivalent? Use MATLAB to check your
answer for speci c values of a, b, c, and d.
a. 1. (a == b) & ((b == c)|(a == c))
2. (a == b)|((b == c)&(a == c))
b. 1. (a < b) & ((a > c)|(a > d))
2. (a < b) & (a > c)|((a < b)&(a > d))
16.
Write a script le using conditional statements to evaluate the following
function, assuming that the scalar variable x has a value. The function is
y e x 1 for x 1, y 2 cos(x) for 1 x 5, and y 10(x 5) 1
for x 5. Use your le to evaluate y for x 5, x 3, and x 15, and
check the results by hand.
Section 4.4
17.
Rewrite the following statements to use only one if statement.
if x < y
if z < 10
w = x*y*z
end
end
18.
Write a program that accepts a numerical value x from 0 to 100 as input
and computes and displays the corresponding letter grade given by the
following table.
A x Ú 90
B 80 … x … 89
C 70 … x … 79
D 60 … x … 69
F x 6 60
a. Use nested if statements in your program (do not use elseif).
b. Use only elseif clauses in your program.
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19.
Write a program that accepts a year and determines whether the year is a
leap year. Use the mod function. The output should be the variable
extra_day, which should be 1 if the year is a leap year and 0 otherwise. The rules for determining leap years in the Gregorian calendar are
as follows:
1. All years evenly divisible by 400 are leap years.
2. Years evenly divisible by 100 but not by 400 are not leap years.
3. Years divisible by 4 but not by 100 are leap years.
4. All other years are not leap years.
20.
For example, the years 1800, 1900, 2100, 2300, and 2500 are not leap
years, but 2400 is a leap year.
Figure P20 shows a mass-spring model of the type used to design packaging systems and vehicle suspensions, for example. The springs exert a
force that is proportional to their compression, and the proportionality
constant is the spring constant k. The two side springs provide additional
resistance if the weight W is too heavy for the center spring. When the
weight W is gently placed, it moves through a distance x before coming to
rest. From statics, the weight force must balance the spring forces at this
new position. Thus
W = k1x
W = k1x + 2k2(x - d)
if x d
if x d
These relations can be used to generate the plot of x versus W.
a. Create a function le that computes the distance x, using the input
parameters W, k1, k2, and d. Test your function for the following
two cases, using the values k1 104 N/m; k2 1.5 104 N/m;
d 0.1 m.
W 500 N
W 2000 N
b. Use your function to plot x versus W for 0
values of k1, k2, and d given in part a.
W
d
3000 N for the
W
Platform
k1
x
k2
Figure P20
k2
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Problems
Section 4.5
21.
Use a for loop to plot the function given in Problem 16 over the interval
2 x 6. Properly label the plot. The variable y represents height in
kilometers, and the variable x represents time in seconds.
22.
Use a for loop to determine the sum of the rst 10 terms in the series
5k3, k 1, 2, 3, . . . , 10.
23.
The (x, y) coordinates of a certain object as a function of time t are given
by
y(t) = 25t2 - 120t + 144
x(t) = 5t - 10
for 0 t 4. Write a program to determine the time at which the object
is the closest to the origin at (0, 0). Determine also the minimum distance.
Do this in two ways:
a. By using a for loop.
b. By not using a for loop.
24.
Consider the array A.
A =
3
-8
J
- 17
5
-1
6
-4
33
K
-9
Write a program that computes the array B by computing the natural logarithm of all the elements of A whose value is no less than 1, and adding
20 to each element that is equal to or greater than 1. Do this in two ways:
a. By using a for loop with conditional statements.
b. By using a logical array as a mask.
25.
We want to analyze the mass-spring system discussed in Problem 20 for
the case in which the weight W is dropped onto the platform attached to
the center spring. If the weight is dropped from a height h above the platform, we can nd the maximum spring compression x by equating the
weight’s gravitational potential energy W(h + x) with the potential
energy stored in the springs. Thus
W (h + x) = 12k1x2
if x < d
which can be solved for x as
x =
W 1W2 + 2k1Wh
k1
if x 6 d
and
W(h + x) = 12 k1x2 + 12(2k2)(x - d)2
if x Ú d
205
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which gives the following quadratic equation to solve for x:
(k1 + 2k2)x2 - (4k2d + 2W)x + 2k2d2 - 2 Wh = 0
if x Ú d
a. Create a function le that computes the maximum compression x due
to the falling weight. The function’s input parameters are k1, k2, d, W,
and h. Test your function for the following two cases, using the
values k1 = 104 N/m; k2 = 1.5 * 104 N/m; and d = 0.1 m.
W = 100 N
h = 0.5 m
W = 2000 N
h = 0.5 m
b. Use your function le to generate a plot of x versus h for 0 h
Use W = 100 N and the preceding values for k1, k2, and d.
26.
2 m.
Electrical resistors are said to be connected “in series” if the same current
passes through each and “in parallel” if the same voltage is applied across
each. If in series, they are equivalent to a single resistor whose resistance
is given by
R = R1 + R2 + R3 + Á + Rn
If in parallel, their equivalent resistance is given by
1
1
1
1
1
=
+
+
+ Á +
R
R1
R2
R3
Rn
Write an M- le that prompts the user for the type of connection (series or
parallel) and the number of resistors n and then computes the equivalent
resistance.
27.
a. An ideal diode blocks the ow of current in the direction opposite that
of the diode’s arrow symbol. It can be used to make a half-wave rectier as shown in Figure P27a. For the ideal diode, the voltage ␷L across
the load RL is given by
␷L = e
␷S
0
if ␷S 7 0
if ␷S … 0
Suppose the supply voltage is
␷S(t) = 3e - t/3 sin(t)
V
where time t is in seconds. Write a MATLAB program to plot the voltage ␷L versus t for 0 … t … 10.
b. A more accurate model of the diode’s behavior is given by the offset
diode model, which accounts for the offset voltage inherent in semiconductor diodes. The offset model contains an ideal diode and a battery whose voltage equals the offset voltage (which is approximately
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Problems
+
Diode
vS
vL
–
(a)
0 .6 V
+
–
+
vL
vS
–
(b)
Figure P27
0.6 V for silicon diodes) [Rizzoni, 2007]. The half-wave recti er using
this model is shown in Figure P27b. For this circuit,
␷ - 0.6
␷L = e S
0
if ␷S 7 0.6
if ␷S … 0.6
Using the same supply voltage given in part a, plot the voltage yL versus
t for 0 … t … 10; then compare the results with the plot obtained in
part a.
28.* A company wants to locate a distribution center that will serve six of its
major customers in a 30 * 30 mi area. The locations of the customers
relative to the southwest corner of the area are given in the following
table in terms of (x, y) coordinates (the x direction is east; the y direction
is north) (see Figure P28). Also given is the volume in tons per week that
must be delivered from the distribution center to each customer. The
weekly delivery cost ci for customer i depends on the volume Vi and the
distance di from the distribution center. For simplicity we will assume that
this distance is the straight-line distance. (This assumes that the road network is dense.) The weekly cost is given by ci = 0.5diVi, i = 1, Á , 6.
Find the location of the distribution center (to the nearest mile) that minimizes the total weekly cost to service all six customers.
207
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CHAPTER 4 Programming with MATLAB
N orth
30
1
20
2
y
(m iles )
3
5
10
6
4
0
East
0
10
20
30
x (m iles)
Figure P28
29.
Customer
x location
(mi)
y location
(mi)
Volume
(tons/week)
1
2
3
4
5
6
1
7
8
17
22
27
28
18
16
2
10
8
3
7
4
5
2
6
A company has the choice of producing up to four different products with
its machinery, which consists of lathes, grinders, and milling machines.
The number of hours on each machine required to produce a product is
given in the following table, along with the number of hours available
per week on each type of machine. Assume that the company can sell
everything it produces. The pro t per item for each product appears in the
last line of the table.
Product
Hours required
Lathe
Grinder
Milling
Unit pro t ($)
1
2
3
4
Hours available
1
0
3
100
2
2
1
150
0.5
4
5
90
3
1
2
120
40
30
45
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Problems
30.
a. Determine how many units of each product the company should
make to maximize its total pro t, and then compute this pro t.
Remember, the company cannot make fractional units, so your
answer must be in integers. (Hint: First estimate the upper limits on
the number of products that can be produced without exceeding the
available capacity.)
b. How sensitive is your answer? How much does the pro t decrease if
you make one more or one less item than the optimum?
A certain company makes televisions, stereo units, and speakers. Its parts
inventory includes chassis, picture tubes, speaker cones, power supplies,
and electronics. The inventory, required components, and pro t for each
product appear in the following table. Determine how many of each product to make in order to maximize the pro t.
Product
Television Stereo unit
Requirements
Chassis
Picture tube
Speaker cone
Power supply
Electronics
Unit pro t ($)
1
1
2
1
2
80
1
0
2
1
2
50
Speaker unit Inventory
0
0
1
0
1
40
450
250
800
450
600
Section 4.6
31.
32.
33.
Plot the function y 10(1 ex/4) over the interval 0 x xmax, using
a while loop to determine the value of xmax such that y(xmax) 9.8.
Properly label the plot. The variable y represents force in newtons, and
the variable x represents time in seconds.
Use a while loop to determine how many terms in the series 2k, k 1,
2, 3, . . . , are required for the sum of the terms to exceed 2000. What is
the sum for this number of terms?
One bank pays 5.5 percent annual interest, while a second bank pays
4.5 percent annual interest. Determine how much longer it will take to
accumulate at least $50 000 in the second bank account if you deposit
$1000 initially and $1000 at the end of each year.
34.* Use a loop in MATLAB to determine how long it will take to accumulate
$1 000 000 in a bank account if you deposit $10 000 initially and $10 000
at the end of each year; the account pays 6 percent annual interest.
35.
A weight W is supported by two cables anchored a distance D apart
(see Figure P35). The cable length LAB is given, but the length LAC is to be
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D
A
C
LAC
LA B
B
W
Figure P35
selected. Each cable can support a maximum tension force equal to W.
For the weight to remain stationary, the total horizontal force and total
vertical force must each be zero. This principle gives the equations
- TAB cos + TAC cos = 0
TAB sin + TAC sin = W
We can solve these equations for the tension forces TAB and TAC if we
know the angles and . From the law of cosines
= cos - 1 a
D2 + L2AB - L2AC
b
2DLAB
From the law of sines
= sin - 1 a
LAB sin b
LAC
For the given values D = 6 ft, LAB = 3 ft, and W = 2000 lb, use a loop
in MATLAB to nd LACmin, the shortest length LAC we can use without
TAB or TAC exceeding 2000 lb. Note that the largest LAC can be is 6.7 ft
(which corresponds to = 90°). Plot the tension forces TAB and TAC on
the same graph versus LAC for LACmin … LAC … 6.7.
36.* In the structure in Figure P36a, six wires support three beams. Wires 1
and 2 can support no more than 1200 N each, wires 3 and 4 can support
no more than 400 N each, and wires 5 and 6 can support no more than
200 N each. Three equal weights W are attached at the points shown.
Assuming that the structure is stationary and that the weights of the wires
and the beams are very small compared to W, the principles of statics applied to a particular beam state that the sum of vertical forces is zero and
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Problems
1
1m
1m
1m
1m
1m
3
1m
1m
2
4
6
W
5
W
(a)
W
T1
1
3
1
T3
T4
1
1
1
W
1
T2
T6
1
(b)
W
T6
T5
1
2
W
Figure P36
that the sum of moments about any point is also zero. Applying these
principles to each beam using the free-body diagrams shown in
Figure P36b, we obtain the following equations. Let the tension force in
wire i be Ti. For beam 1
T1 + T2 = T3 + T4 + W + T6
-T3 - 4T4 - 5W - 6T6 + 7T2 = 0
For beam 2
T3 + T4 = W + T5
- W - 2T5 + 3T4 = 0
For beam 3
T5 + T6 = W
- W + 3T6 = 0
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R1
i1
i2
R2
i3
i4
R3
i5
+
+
v1
R4
R5
–
v2
–
Figure P37
Find the maximum value of the weight W the structure can support.
Remember that the wires cannot support compression, so Ti must be
nonnegative.
37.
The equations describing the circuit shown in Figure P37 are
- y1 + R1i1 + R4i4 = 0
- R4i4 + R2i2 + R5i5 = 0
- R5i5 + R3i3 + y2 = 0
i1 = i2 + i4
i2 = i3 + i5
a. The given values of the resistances and the voltage y1 are R1 = 5,
R2 = 100, R3 = 200, R4 = 150, R5 = 250 k, and y1 = 100 V.
(Note that 1 k 1000 .) Suppose that each resistance is rated to
carry a current of no more than 1 mA ( 0.001 A). Determine the
allowable range of positive values for the voltage y2.
b. Suppose we want to investigate how the resistance R3 limits the
allowable range for y2. Obtain a plot of the allowable limit on y2 as a
function of R3 for 150 … R3 … 250 k.
38.
Many applications require us to know the temperature distribution in an
object. For example, this information is important for controlling the material properties, such as hardness, when cooling an object formed from
molten metal. In a heat-transfer course, the following description of the
temperature distribution in a at, rectangular metal plate is often derived.
The temperature is held constant at T1 on three sides and at T2 on the
fourth side (see Figure P38). The temperature T(x, y) as a function of the
xy coordinates shown is given by
T (x, y) = (T2 - T1)w(x, y) + T1
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Problems
y
T2
W
T ( x ,y )
T1
T1
0
0
T1
L
x
Figure P38
where
w(x, y) =
q
2
2
nx sinh(ny/L)
b
sin a
a
n odd n
L sinh(nW/L)
Use the following data: T1 = 70°F, T2 = 200°F, and W = L = 2 ft.
a. The terms in the preceding series become smaller in magnitude as
n increases. Write a MATLAB program to verify this fact for
n = 1, Á , 19 for the center of the plate (x = y = 1).
b. Using x = y = 1, write a MATLAB program to determine how many
terms are required in the series to produce a temperature calculation
that is accurate to within 1 percent. (That is, for what value of n will
the addition of the next term in the series produce a change in T of less
than 1 percent?) Use your physical insight to determine whether this
answer gives the correct temperature at the center of the plate.
c. Modify the program from part b to compute the temperatures in the
plate; use a spacing of 0.2 for both x and y.
39.
Consider the following script le. Fill in the lines of the following table
with the values that would be displayed immediately after the while
statement if you ran the script le. Write in the values the variables have
each time the while statement is executed. You might need more or
fewer lines in the table. Then type in the le, and run it to check your
answers.
k = 1;b = -2;x = -1;y = -2;
while k <= 3
k, b, x, y
y = x^2 - 3;
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if y < b
b = y;
end
x = x + 1;
k = k + 1;
end
Pass
k
b
x
y
First
Second
Third
Fourth
Fifth
40.
Assume that the human player makes the rst move against the computer
in a game of Tic-Tac-Toe, which has a 3 3 grid. Write a MATLAB
function that lets the computer respond to that move. The function’s input
argument should be the cell location of the human player’s move. The
function’s output should be the cell location of the computer’s rst move.
Label the cells as 1, 2, 3 across the top row; 4, 5, 6 across the middle row;
and 7, 8, 9 across the bottom row.
Section 4.7
41.
The following table gives the approximate values of the static coef cient
of friction for various materials.
Materials
␮
Metal on metal
Wood on wood
Metal on wood
Rubber on concrete
0.20
0.35
0.40
0.70
To start a weight W moving on a horizontal surface, you must push with
a force F, where F = W. Write a MATLAB program that uses the
switch structure to compute the force F. The program should accept as
input the value of W and the type of materials.
42.
The height and speed of a projectile (such as a thrown ball) launched with
a speed of y0 at an angle A to the horizontal are given by
h(t) = ␷0t sin A - 0.5gt2
␷(t) = 1 ␷02 - 2␷0gt sin A + g2t2
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Problems
where g is the acceleration due to gravity. The projectile will strike the
ground when h(t) = 0, which gives the time to hit thit = 2(␷0 >g) sin A.
Use the switch structure to write a MATLAB program to compute
the maximum height reached by the projectile, the total horizontal distance
traveled, or the time to hit. The program should accept as input the user’s
choice of which quantity to compute and the values of ␷0, A, and g. Test
the program for the case where y0 = 40 m/s, A = 300, and g = 9.81 m/s2.
43.
Use the switch structure to write a MATLAB program to compute the
amount of money that accumulates in a savings account in one year. The
program should accept the following input: the initial amount of money
deposited in the account; the frequency of interest compounding
(monthly, quarterly, semiannually, or annually); and the interest rate.
Run your program for a $1000 initial deposit for each case; use a 5 percent interest rate. Compare the amounts of money that accumulate for
each case.
44.
Engineers often need to estimate the pressures and volumes of a gas in a
container. The van der Waals equation is often used for this purpose. It is
P =
RT
a
- 2
N
V - b
VN
where the term b is a correction for the volume of the molecules and the
term a/VN 2 is a correction for molecular attractions. The gas constant is R,
the absolute temperature is T, and the gas speci c volume is VN . The value
of R is the same for all gases; it is R 0.08206 L-atm/mol-K. The values
of a and b depend on the type of gas. Some values are given in the following table. Write a user-de ned function using the switch structure that
computes the pressure P on the basis of the van der Waals equation. The
function’s input arguments should be T, VN , and a string variable containing the name of a gas listed in the table. Test your function for chlorine
(Cl2) for T 300 K and VN 20 L/mol.
Gas
Helium, He
Hydrogen, H2
Oxygen, O2
Chlorine, Cl2
Carbon dioxide, CO2
45.
a (L2-atm/mol2)
0.0341
0.244
1.36
6.49
3.59
b (L/mol)
0.0237
0.0266
0.0318
0.0562
0.0427
Using the program developed in Problem 19, write a program that uses
the switch structure to compute the number of days in a year up to a
given date, given the year, the month, and the day of the month.
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Section 4.9
46.
Consider the college enrollment model discussed in Example 4.9–2. Suppose the college wants to limit freshman admissions to 120 percent of the
current sophomore class and limit sophomore transfers to 10 percent of
the current freshman class. Rewrite and run the program given in the
example to examine the effects of these policies over a 10-year period.
Plot the results.
47.
Suppose you project that you will be able to deposit the following
monthly amounts into a savings account for a period of 5 years. The
account initially has no money in it.
Year
Monthly deposit ($)
1
300
2
350
3
350
4
350
5
400
At the end of each year in which the account balance is at least $3000,
you withdraw $2000 to buy a certi cate of deposit (CD), which pays
6 percent interest compounded annually.
Write a MATLAB program to compute how much money will
accumulate in 5 years in the account and in any CDs you buy. Run
the program for two different savings interest rates: 4 percent and
5 percent.
48.* A certain company manufactures and sells golf carts. At the end of each
week, the company transfers the carts produced that week into storage
(inventory). All carts that are sold are taken from the inventory. A simple
model of this process is
I(k + 1) = P(k) + I(k) - S(k)
where
P(k) = number of carts produced in week k
I(k) = number of carts in inventory in week k
S(k) = number of carts sold in week k
The projected weekly sales for 10 weeks are
Week
Sales
1
50
2
55
3
60
4
70
5
70
6
75
7
80
8
80
9 10
90 55
Suppose the weekly production is based on the previous week’s sales so
that P(k) = S(k - 1). Assume that the rst week's production is 50 carts;
that is, P(1) = 50. Write a MATLAB program to compute and plot the
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Problems
49.
number of carts in inventory for each of the 10 weeks or until the inventory drops below zero. Run the program for two cases: (a) an initial
inventory of 50 carts so that I(1) = 50 and (b) an initial inventory of
30 carts so that I(1) = 30.
Redo Problem 48 with the restriction that the next week’s production is
set to zero if the inventory exceeds 40 carts.
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Courtesy of Aero Vironment, Inc.
Engineering in the
21st Century. . .
Low-Speed Aeronautics
S
ometimes just when we think a certain technical area is mature and the
possibility of further development is unlikely, we are surprised by a novel
design. Recent developments in low-speed aeronautics are examples of this
phenomenon. Even though engineers have known for years that a human could
generate enough power to propel an aircraft, the feat remained impossible until
the availability of lightweight materials that enabled the Gossamer Challenger to
y across the English Channel. Solar -powered aircraft that can stay aloft for over
a day are other examples.
Another example is the recent appearance of wing-in-ground (WIG) effect
vehicles. WIG vehicles make use of an air cushion to create lift. They are a hybrid
between an aircraft and a hovercraft, and most are intended for over-water ight
only. A hovercraft rides on an air cushion created by fans, but the air cushion of
a WIG vehicle is due to the air that is captured under its stubby wings.
Small aircraft with cameras will be useful for search and reconnaissance.
An example of such a “micro air vehicle” (MAV) is the 6-in.-long Black Widow
produced by Aero Vironment, Inc. It carries a 2-g video camera the size of a
sugar cube, and ies at about 65 km/h with a range of 10 km. Proper design of
such vehicles requires a systematic methodology to nd the optimum combination of airfoil shape, motor type, battery type, and most importantly, the propeller
shape.
The MATLAB advanced graphics capabilities make it useful for visualizing flow patterns, and the Optimization toolbox is useful for designing such
vehicles. ■
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C H A P T E R
5
Advanced Plotting
OUTLINE
5.1 xy Plotting Functions
5.2 Additional Commands and Plot Types
5.3 Interactive Plotting in MATLAB
5.4 Three-Dimensional Plots
5.5 Summary
Problems
In this chapter you will learn additional features to use to create a variety of
two-dimensional plots, which are also called xy plots, and three-dimensional
plots called xyz plots, or surface plots. Two-dimensional plots are discussed in
Sections 5.1 through 5.3. Section 5.4 discusses three-dimensional plots. These
plotting functions are described in the graph2d and graph3d Help categories, so typing help graph2d or help graph3d will display a list of
the relevant plotting functions.
An important application of plotting is function discovery, the technique for
using data plots to obtain a mathematical function or “mathematical model” that
describes the process that generated the data. This topic is treated in Chapter 6.
5.1 xy Plotting Functions
The “anatomy” and nomenclature of a typical xy plot is shown in Figure 5.1–1,
in which the plot of a data set and a curve generated from an equation appear. A
plot can be made from measured data or from an equation. When data are plotted, each data point is plotted with a data symbol, or point marker, such as the
small circles shown in Figure 5.1–1. An exception to this rule would be when
there are so many data points that the symbols would be too densely packed.
DATA SYMBOL
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PLOT TITLE
Height of a Falling Object Versus Time
1600
DATA SYMBOL
1400
1200
Height (feet)
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1000
LEGEND
800
600
Zero Drag Model
Data
400
TICK MARK
200
0
0
1
2
3
4
5
6
Time (seconds)
7
8
9
10
TICK- MARK LABEL
AXIS LABEL
Figure 5.1–1 Nomenclature for a typical xy plot.
In that case, the data points should be plotted with a dot. However, when the plot
is generated from a function, data symbols must never be used! Lines between
closely spaced points are always used to plot a function.
The MATLAB basic xy plotting function is plot(x,y) as we saw in
Chapter 1. If x and y are vectors, a single curve is plotted with the x values on
the abscissa and the y values on the ordinate. The xlabel and ylabel commands put labels on the abscissa and the ordinate, respectively. The syntax is
xlabel(‘text’), where text is the text of the label. Note that you must
enclose the label’s text in single quotes. The syntax for ylabel is the same.
The title command puts a title at the top of the plot. Its syntax is
title(‘text’), where text is the title’s text.
The plot(x,y) function in MATLAB automatically selects a tick-mark
spacing for each axis and places appropriate tick labels. This feature is called
autoscaling. MATLAB also chooses limits for the x and y axes. The order of the
xlabel, ylabel, and title commands does not matter, but we must place
them after the plot command, either on separate lines using ellipses or on the
same line separated by commas.
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5.1 xy Plotting Functions
After the plot command is executed, the plot will appear in the Figure
window. You can obtain a hard copy of the plot in one of several ways:
1. Use the menu system. Select Print on the File menu in the Figure window.
Answer OK when you are prompted to continue the printing process.
2. Type print at the command line. This command sends the current plot
directly to the printer.
3. Save the plot to a le to be printed later or imported into another application such as a word processor. You need to know something about graphics
le formats to use this le properly . See the subsection Exporting Figures
later in this section.
Type help print to obtain more information.
MATLAB assigns the output of the plot command to Figure window
number 1. When another plot command is executed, MATLAB overwrites
the contents of the existing Figure window with the new plot. Although you
can keep more than one Figure window active, we do not use this feature in this
text.
When you have nished with the plot, close the Figure window by selecting
Close from the File menu in the Figure window. If you do not close the window,
it will not reappear when a new plot command is executed. However, the
g ure will still be updated.
Table 5.1–1 lists the requirements essential to producing plots that communicate effectively.
Table 5.1–1 Requirements for a correct plot
1.
2.
3.
4.
5.
6.
7.
Each axis must be labeled with the name of the quantity being plotted and its units! If
two or more quantities having different units are plotted (such as when in a plot of both
speed and distance versus time), indicate the units in the axis label, if there is room, or
in the legend or labels for each curve.
Each axis should have regularly spaced tick marks at convenient intervals—not too
sparse, but not too dense—with a spacing that is easy to interpret and interpolate. For
example, use 0.1, 0.2, and so on, rather than 0.13, 0.26, and so on.
If you are plotting more than one curve or data set, label each on its plot, use different
line types, or use a legend to distinguish them.
If you are preparing multiple plots of a similar type or if the axes’ labels cannot convey
enough information, use a title.
If you are plotting measured data, plot each data point with a symbol such as a circle,
square, or cross (use the same symbol for every point in the same data set). If there are
many data points, plot them using the dot symbol.
Sometimes data symbols are connected by lines to help the viewer visualize the data,
especially if there are few data points. However, connecting the data points, especially
with a solid line, might be interpreted to imply knowledge of what occurs between the
data points. Thus you should be careful to prevent such misinterpretation.
If you are plotting points generated by evaluating a function (as opposed to measured
data), do not use a symbol to plot the points. Instead, be sure to generate many points,
and connect the points with solid lines.
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grid and axis Commands
AXIS LIMITS
The grid command displays gridlines at the tick marks corresponding to the
tick labels. You can use the axis command to override the MATLAB selections
for the axis limits. The basic syntax is axis([xmin xmax ymin ymax]).
This command sets the scaling for the x and y axes to the minimum and maximum values indicated. Note that, unlike an array, this command does not use
commas to separate the values.
Figure 5.1–2 shows a plot in which the command axis([0 10 2 5])
was used to override the limits chosen by auto scaling (which chose the upper
limit of the ordinate to be 4).
The axis command has the following variants:
■
■
■
axis square selects the axes’ limits so that the plot will be square.
axis equal selects the scale factors and tick spacing to be the same on
each axis. This variation makes plot(sin(x),cos(x)) look like a
circle, instead of an oval.
axis auto returns the axis scaling to its default autoscaling mode in
which the best axes’ limits are computed automatically.
Type help axis to see the full list of variants.
Figure 5.1–2 A sample plot shown in a Figure window.
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5.1 xy Plotting Functions
Plots of Complex Numbers
With only one argument, say, plot(y), the plot function will plot the values
in the vector y versus their indices 1, 2, 3, . . . , and so on. If y is complex,
plot(y) plots the imaginary parts versus the real parts. Thus plot(y) in this
case is equivalent to plot(real(y),imag(y)). This situation is the only
time when the plot function handles the imaginary parts; in all other variants of
the plot function, it ignores the imaginary parts. For example, the script le
z = 0.1 + 0.9i;
n = 0:0.01:10;
plot(z.^n),xlabel(‘Real’),ylabel(‘Imaginary’)
generates a spiral plot.
The Function Plot Command fplot
MATLAB has a “smart” command for plotting functions. The fplot command
automatically analyzes the function to be plotted and decides how many plotting
points to use so that the plot will show all the features of the function. Its syntax
is fplot(function, [xmin xmax]), where function is a function handle to the function to be plotted and [xmin xmax] speci es the minimum and
maximum values of the independent variable. The range of the dependent
variable can also be speci ed. In this case the syntax is fplot(function,
[xmin xmax ymin ymax]).
For example, the session
>>f = @(x) (cos(tan(x)) - tan(sin(x)));
>>fplot(f,[1 2])
produces the plot shown in Figure 5.1–3a. The fplot command automatically
chooses enough plotting points to display all the variations in the function. We
can achieve the same results using the plot command, but we need to know
how many values to compute to generate the plot. For example, choosing a spacing of 0.01, and using plot, we obtain the plot in Figure 5.1–3b. We see that this
choice of spacing misses some of the function’s behavior.
Another form is [x,y] = fplot(function, limits), where limits
may be either [xmin xmax] or [xmin xmax ymin ymax]. With this form
the command returns the abscissa and ordinate values in the column vectors x
and y, but no plot is produced. The returned values can then be used for other
purposes, such as plotting multiple curves, which is the topic of the next section.
Other commands can be used with the fplot command to enhance a plot’s appearance, for example, the title, xlabel, and ylabel commands and the
line type commands to be introduced in the next section.
Plotting Polynomials
We can plot polynomials more easily by using the polyval function. The
function polyval(p,x) evaluates the polynomial p at speci ed values of its
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−0.5
−0.5
−1
−1
−1.5
−1.5
−2
−2
−2.5
−2.5
−3
1.3
1.4
1.5
1.6
1.7
(a)
1.8
−3
1.3
1.4
1.5
1.6
1.7
1.8
(b)
Figure 5.1–3 (a) The plot was generated with fplot. (b) The plot was generated with
plot using 101 points.
independent variable x. For example, to plot the polynomial 3x5 2x4 100x3 2x2 7x 90 over the range 6 x 6 with a spacing of 0.01, you type
>>x = -6:0.01:6;
>>p = [3,2,-100,2,-7,90];
>>plot(x,polyval(p,x)),xlabel(‘x’),ylabel(‘p’)
Table 5.1–2 summarizes the xy plotting commands discussed in this section.
Test Your Understanding
T5.1–1 Plot the equation y = 0.4 11.8x for 0 x 35 and 0 y 3.5.
T5.1–2 Use the fplot command to investigate the function tan(cos x) sin(tan x) for 0 x 2. How many values of x are needed to obtain
the same plot using the plot command? (Answer: 292 values.)
T5.1–3 Plot the imaginary part versus the real part of the function (0.2 0.8i)n
for 0 n 20. Choose enough points to obtain a smooth curve. Label
each axis and put a title on the plot. Use the axis command to
change the tick-label spacing.
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5.1 xy Plotting Functions
225
Table 5.1–2 Basic xy plotting commands
Command
Description
axis([xmin xmax ymin ymax])
fplot(function,[xmin xmax])
Sets the minimum and maximum limits of the x and y axes.
Performs intelligent plotting of functions, where function is a function handle that describes the function to be plotted and [xmin xmax]
speci es the minimum and maximum values of the independent variable. The range of the dependent variable can also be speci ed. In this
case the syntax is fplot(function, [xmin xmax ymin
ymax]).
Displays gridlines at the tick marks corresponding to the tick labels.
Generates a plot of the array y versus the array x on rectilinear axes.
Plots the values of y versus their indices if y is a vector. Plots the
imaginary parts of y versus the real parts if y is a vector having
complex values.
Evaluates the polynomial p at speci ed values of its independent
variable x.
Prints the plot in the Figure window.
Puts text in a title at the top of a plot.
Adds a text label to the x axis (the abscissa).
Adds a text label to the y axis (the ordinate).
grid
plot(x,y)
plot(y)
polyval(p,x)
print
title(‘text’)
xlabel(‘text’)
ylabel(‘text’)
Saving Figures
When you create a plot, the Figure window appears. This window has eight
menus, which are discussed in detail in Section 5.3. The File menu is used for
saving and printing the gure. You can save your gure in a format that can be
opened during another MATLAB session or in a format that can be used by other
applications.
To save a gure that can be opened in subsequent MA TLAB sessions, save
it in a gure le with the . g le name extension.
To do this, select Save from the
Figure window File menu or click the Save button (the disk icon) on the toolbar.
If this is the rst time you are saving the le, the Save As dialog box appears.
Make sure that the type is MATLAB Figure (*. g). Specify the name you want
assigned to the gure le. Click OK. You can also use the saveas command.
To open a gure le, select Open from the File menu or click the Open button (the opened folder icon) on the toolbar. Select the gure le you want to open
and click OK. The gure le appears in a new gure window .
Exporting Figures
If you want to save the gure in a format that can be used by another application,
such as the standard graphics le formats TIFF or EPS, perform these steps.
1. Select Export Setup from the File menu. This dialog provides options you
can specify for the output le, such as the gure size, fonts, line size and
style, and output format.
2. Select Export from the Export Setup dialog. A standard Save As dialog
appears.
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Table 5.1–3 Hints for improving plots
1.
2.
3.
4.
Start scales from zero whenever possible. This technique prevents a false impression
of the magnitudes of any variations shown on the plot.
Use sensible tick-mark spacing. For example, if the quantities are months, choose a
spacing of 12 because 1兾10 of a year is not a convenient division. Space tick marks
as close as is useful, but no closer.
Minimize the number of zeros in the tick labels. For example, use a scale in millions
of dollars when appropriate, instead of a scale in dollars with six zeros after every
number.
Determine the minimum and maximum data values for each axis before plotting the
data. Then set the axis limits to cover the entire data range plus an additional amount
to allow convenient tick-mark spacing to be selected.
3. Select the format from the list of formats in the Save As type menu. This
selects the format of the exported le and adds the standard le name
extension given to les of that type.
4. Enter the name you want to give the le, less the extension.
5. Click Save.
You can also export the gure from the command line, by using the print
command. See MATLAB Help for more information about exporting gures in
different formats.
You can also copy a gure to the clipboard and then paste it into another
application:
1. Select Copy Options from the Edit menu of the Figure window. The
Copying Options page of the Preferences dialog box appears.
2. Complete the elds on the Copying Options page and click OK.
3. Select Copy Figure from the Edit menu.
The gure is copied to the Windows clipboard and can be pasted into another
application.
MATLAB also enables you to save gures in formats compatible with
PowerPoint and MS Word. See the MATLAB Help for more information.
The graphics functions covered in this section and in Section 5.3 can be
placed in script les that can be reused to create similar plots. This feature gives
them an advantage over the interactive plotting tools discussed in Section 5.3.
When you are creating plots, keep in mind that the actions listed in
Table 5.1–3, while not required, can never the less improve the appearance and
usefulness of your plots.
5.2 Additional Commands and Plot Types
SUBPLOT
MATLAB can create gures that contain an array of plots, called subplots. These
are useful when you want to compare the same data plotted with different axis
types, for example. The MATLAB subplot command creates such gures. We
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frequently need to plot more than one curve or data set on a single plot. Such a
plot is called an overlay plot. This section describes these plots and several other
types of plots.
Subplots
You can use the subplot command to obtain several smaller “subplots” in the
same gure. The syntax is subplot(m,n,p). This command divides the
Figure window into an array of rectangular panes with m rows and n columns.
The variable p tells MATLAB to place the output of the plot command following the subplot command into the pth pane. For example, subplot(3,2,5)
creates an array of six panes, three panes deep and two panes across, and directs
the next plot to appear in the fth pane (in the bottom left corner). The following
script le created Figure 5.2–1, which shows the plots of the functions y e1.2x sin(10x 5) for 0 x 5 and y |x3 100| for 6 x 6.
x = 0:0.01:5;
y = exp(-1.2*x).*sin(10*x+5);
subplot(1,2,1)
plot(x,y),xlabel(‘x’),ylabel(‘y’),axis([0 5 -1 1])
x = -6:0.01:6;
y = abs(x.^3-100);
subplot(1,2,2)
plot(x,y),xlabel(‘x’),ylabel(‘y’),axis([-6 6 0 350])
1
350
0.8
300
0.6
250
0.4
200
0
y
y
0.2
150
–0.2
–0.4
100
–0.6
50
–0.8
–1
0
1
2
x
3
4
0
–5
Figure 5.2–1 Application of the subplot command.
0
x
5
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Test Your Understanding
T5.2–1 Pick a suitable spacing for t and ␷, and use the subplot command to
plot the function z e0.5t cos(20t 6) for 0 t 8 and the function u 6 log10 (␷ 2 20) for 8 ␷ 8. Label each axis.
Overlay Plots
You can use the following variants of the MATLAB basic plotting functions
plot(x,y) and plot(y) to create overlay plots:
■
■
■
■
plot(A) plots the columns of A versus their indices and generates n
curves, where A is a matrix with m rows and n columns.
plot(x,A) plots the matrix A versus the vector x, where x is either a
row vector or a column vector and A is a matrix with m rows and n columns.
If the length of x is m, then each column of A is plotted versus the vector x.
There will be as many curves as there are columns of A. If x has length n,
then each row of A is plotted versus the vector x. There will be as many
curves as there are rows of A.
plot(A,x) plots the vector x versus the matrix A. If the length of x is m,
then x is plotted versus the columns of A. There will be as many curves as
there are columns of A. If the length of x is n, then x is plotted versus the
rows of A. There will be as many curves as there are rows of A.
plot(A,B) plots the columns of the matrix B versus the columns of the
matrix A.
Data Markers and Line Types
To plot the vector y versus the vector x and mark each point with a data
marker, enclose the symbol for the marker in single quotes in the plot function.
Table 5.2–1 shows the symbols for some of the available data markers. For
example, to use a small circle, which is represented by the lowercase letter o,
type plot(x,y, ‘o’). This notation results in a plot like the one on the left
Table 5.2–1 Speci ers for data markers, line types, and colors
Data markers†
Dot ( # )
Asterisk (*)
Cross ()
Circle (o)
Plus sign ()
Square ( )
Diamond (䉫)
Five-pointed star (夹)
†
Line types
#
*
o
s
d
p
Solid line
Dashed line
Dash-dotted line
Dotted line
Colors
--.
:
Other data markers are available. Search for “markers” in MATLAB Help.
Black
Blue
Cyan
Green
Magenta
Red
White
Yellow
k
b
c
g
m
r
w
y
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3
3
2.5
2.5
2
2
y
3.5
y
3.5
Additional Commands and Plot Types
1.5
1.5
1
1
0.5
0.5
0
0
5
x
10
0
0
5
x
10
Figure 5.2–2 Use of data markers.
in Figure 5.2–2. To connect each data marker with a straight line, we must plot
the data twice, by typing plot(x,y,x,y,‘o’). See the plot on the right in
Figure 5.2–2.
Suppose we have two curves or data sets stored in the vectors x, y, u, and v.
To plot y versus x and v versus u on the same plot, type plot(x,y,u,v). Both
sets will be plotted with a solid line, which is the default line style. To distinguish
the sets, we can plot them with different line types. To plot y versus x with a solid
line and u versus v with a dashed line, type plot(x,y,u,v,’’), where the
symbols ‘’ represent a dashed line. Table 5.2–1 gives the symbols for other line
types. To plot y versus x with asterisks (*) connected with a dotted line, you must
plot the data twice by typing plot(x,y,’*’,x,y,’:’).
You can obtain symbols and lines of different colors by using the color symbols
shown in Table 5.2–1. The color symbol can be combined with the data-marker symbol and the line-type symbol. For example, to plot y versus x with green asterisks (*)
connected with a red dashed line, you must plot the data twice by typing
plot(x,y,’g*’,x,y,’r’). (Do not use colors if you are going to print the
plot on a black-and-white printer.)
Labeling Curves and Data
When more than one curve or data set is plotted on a graph, we must distinguish
between them. If we use different data symbols or different line types, then we
must either provide a legend or place a label next to each curve. To create a legend,
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3.5
Hyperbolic Sine and Hyperbolic Tangent
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sinh(x)
tanh(x)
2.5
2
1.5
1
0.5
0
0
0.2
0.4
0.6
0.8
1
x
1.2
1.4
1.6
1.8
2
Figure 5.2–3 Application of the legend command.
use the legend command. The basic form of this command is legend
(‘string1’,‘string2’), where string1 and string2 are text strings
of your choice. The legend command automatically obtains from the plot the
line type used for each data set and displays a sample of this line type in the legend box next to the string you selected. The following script le produced the
plot in Figure 5.2–3.
x = 0:0.01:2;
y = sinh(x);
z = tanh(x);
plot(x,y,x,z,‘’),xlabel(‘x’),...
ylabel(‘Hyperbolic Sine and Hyperbolic Tangent’),...
legend(‘sinh(x)’,’tanh(x)’)
The legend command must be placed somewhere after the plot command.
When the plot appears in the Figure window, use the mouse to position the legend box. (Hold down the left button on the mouse to move the box.)
Another way to distinguish curves is to place a label next to each. The label
can be generated either with the gtext command, which lets you place the
label by using the mouse, or with the text command, which requires you to
specify the coordinates of the label. The syntax of the gtext command is
gtext(‘string’), where string is a text string that speci es the label of
your choice. When this command is executed, MATLAB waits for a mouse button
or a key to be pressed while the mouse pointer is within the Figure window; the label
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is placed at that position of the mouse pointer. You may use more than one gtext
command for a given plot. The text command text(x,y,‘string’)adds a
text string to the plot at the location speci ed by the coordinates x,y. These coordinates are in the same units as the plot’s data. Of course, nding the proper coordinates to use with the text command usually requires some trial and error.
The hold Command
The hold command creates a plot that needs two or more plot commands.
Suppose we wanted to plot y2 4 ex cos 6x versus y1 3 ex sin 6x, 1 x 1 on the same plot with the complex function z (0.1 0.9i)n, where 0 n 10. The following script le creates the plot in Figure 5.2–4.
x = -1:0.01:1;
y1 = 3+exp(-x).*sin(6*x);
y2 = 4+exp(-x).*cos(6*x);
plot((0.1+0.9i).^(0:0.01:10)),hold,plot(y1,y2),...
gtext(‘y2 versus y1’),gtext(‘Imag(z) versus Real(z)’)
When more than one plot command is used, do not place any of the gtext commands before any plot command. Because the scaling changes as each plot
command is executed, the label placed by the gtext command might end up in the
wrong position. Table 5.2–2 summarizes the plot enhancement commands
introduced in this section.
7
6
y2 versus y1
5
4
3
2
1
Imag (z) versus Real (z)
0
–1
–1
0
1
2
3
Figure 5.2–4 Application of the hold command.
4
5
6
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Table 5.2–2 Plot enhancement commands
Command
Description
gtext(‘text’)
hold
legend(‘leg1’,’leg2’,...)
Places the string text in the Figure window at a point speci ed by the mouse.
Freezes the current plot for subsequent graphics commands.
Creates a legend using the strings leg1, leg2, and so on and enables its
placement with the mouse.
Plots, on rectilinear axes, four arrays: y versus x and v versus u.
Plots the array y versus the array x on rectilinear axes, using the line type,
data marker, and colors speci ed in the string type. See Table 5.2–1.
Plots the columns of the m n array A versus their indices and generates n
curves.
Plots array Q versus array P. See the text for a description of the possible
variants involving vectors and/or matrices: plot(x,A), plot(A,x), and
plot(A,B).
Splits the Figure window into an array of subwindows with m rows and n
columns and directs the subsequent plotting commands to the pth subwindow.
Places the string text in the Figure window at a point speci ed by
coordinates x, y.
plot(x,y,u,v)
plot(x,y,’type’)
plot(A)
plot(P,Q)
subplot(m,n,p)
text(x,y,’text’)
Test Your Understanding
T5.2–2 Plot the following two data sets on the same plot. For each set, x 0, 1,
2, 3, 4, 5. Use a different data marker for each set. Connect the markers
for the rst set with solid lines. Connect the markers for the second set
with dashed lines. Use a legend, and label the plot axes appropriately.
The rst set is y 11, 13, 8, 7, 5, 9. The second set is y 2, 4, 5, 3, 2, 4.
T5.2–3 Plot y cosh x and y 0.5e x on the same plot for 0 x 2. Use different line types and a legend to distinguish the curves. Label the plot axes
appropriately.
T5.2–4 Plot y sinh x and y 0.5e x on the same plot for 0 x 2. Use a solid
line type for each, the gtext command to label the sinh x curve, and the
text command to label the 0.5e x curve. Label the plot axes appropriately.
T5.2–5 Use the hold command and the plot command twice to plot y sin x
and y x x3/3 on the same plot for 0 x 1. Use a solid line type
for each, and use the gtext command to label each curve. Label the
plot axes appropriately.
Annotating Plots
You can create text, titles, and labels that contain mathematical symbols, Greek
letters, and other effects such as italics. The features are based on the TEX
typesetting language. For more information, including a list of the available
characters, search the online Help for the “Text Properties” page. See also the
“Mathematical symbols, Greek Letters, and TEX Characters” page.
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You can create a title having the mathematical function Aet/ sin( t) by
typing
>>title(‘{\it Ae}^{-{\it t/\tau}}\sin({\it \omega t})’)
The backslash character \ precedes all TEX character sequences. Thus the strings
\tau and \omega represent the Greek letters and . Superscripts are created
by typing ^; subscripts are created by typing _. To set multiple characters as superscripts or subscripts, enclose them in braces. For example, type x_{13} to produce x13. In mathematical text variables are usually set in italic, and functions, such
as sin, are set in roman type. To set a character, say, x, in italic using the TEX
commands, you type {\it x}.
Logarithmic Plots
Logarithmic scales—abbreviated log scales—are widely used (1) to represent a
data set that covers a wide range of values and (2) to identify certain trends in data.
Certain types of functional relationships appear as straight lines when plotted using
a log scale. This method makes it easier to identify the function. A log-log plot has
log scales on both axes. A semilog plot has a log scale on only one axis.
Figure 5.2–5 shows a rectilinear plot and a log-log plot of the function
y =
100(1 - 0.01x2)2 + 0.02x2
(1 - x2)2 + 0.1x2
0.1 … x … 100
(5.2–1)
102
35
30
101
25
y
y
20
100
15
10
10−1
5
0
0
50
x
(a)
100
10−2 −2
10
100
x
102
(b)
Figure 5.2–5 (a) Rectilinear plot of the function in Equation (5.2–1). (b) Log-log plot of
the function. Note the wide range of values of both x and y.
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Because of the wide range in values on both the abscissa and the ordinate,
rectilinear scales do not reveal the important features. The following program
produced Figure 5.2–5.
% Create the Rectilinear Plot
x1 = 0:0.01:100; u1 = x1.^2;
num1 = 100*(1-0.01*u1).^2 + 0.02*u1;
den1 = (1-u1).^2 + 0.1*u1;
y1 = sqrt(num1./den1);
subplot(1,2,1), plot(x1,y1),xlabel(‘x’),ylabel(‘y’),
% Create the Loglog Plot
x2 = logspace(-2, 2, 500); u2 = x2.^2;
num2 = 100*(1-0.01*u2).^2 + 0.02*u2;
den2 = (1-u2).^2 + 0.1*u2;
y2 = sqrt(num2./den2);
subplot(1,2,2), loglog(x2,y2),xlabel(‘x’),ylabel(‘y’)
It is important to remember the following points when using log scales:
1. You cannot plot negative numbers on a log scale, because the logarithm of
a negative number is not de ned as a real number .
2. You cannot plot the number 0 on a log scale, because log10 0 ln 0 .
You must choose an appropriately small number as the lower limit on the
plot.
3. The tick-mark labels on a log scale are the actual values being plotted; they
are not the logarithms of the numbers. For example, the range of x values
in the plot in Figure 5.2–5b is from 102 0.01 to 102 100, and the
range of y values is from 102 to 102 100.
MATLAB has three commands for generating plots having log scales. The
appropriate command depends on which axis must have a log scale. Follow these
rules:
1. Use the loglog(x,y) command to have both scales logarithmic.
2. Use the semilogx(x,y) command to have the x scale logarithmic and
the y scale rectilinear.
3. Use the semilogy(x,y) command to have the y scale logarithmic and
the x scale rectilinear.
Table 5.2–3 summarizes these functions. For other two-dimensional plot types,
type help specgraph. We can plot multiple curves with these commands
just as with the plot command. In addition, we can use the other commands, such
as grid, xlabel, and axis, in the same manner. Figure 5.2–6 shows how
these commands are applied. It was created with the following program.
x1 = 0:0.01:3; y1 = 25*exp(0.5*x1);
y2 = 40*(1.7.^x1);
x2 = logspace(-1,1,500); y3 = 15*x2.^(0.37);
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subplot(1,2,1),semilogy(x1,y1,x1,y2, ‘--’),...
legend (‘y = 25e^{0.5x}’, ‘y = 40(1.7) ^x’),...
xlabel(‘x’),ylabel(‘y’),grid,...
subplot(1,2,2),loglog(x2,y3),legend(‘y = 15x^{0.37}’),...
xlabel(‘x’),ylabel(‘y’),grid
Table 5.2–3 Specialized plot commands
Command
Description
bar(x,y)
loglog(x,y)
plotyy(x1,y1,x2,y2)
Creates a bar chart of y versus x.
Produces a log-log plot of y versus x.
Produces a plot with two y axes, y1 on the left and y2
on the right.
Produces a polar plot from the polar coordinates
theta and r, using the line type, data marker, and
colors speci ed in the string type.
Produces a semilog plot of y versus x with logarithmic
abscissa scale.
Produces a semilog plot of y versus x with logarithmic
ordinate scale.
Produces a stairs plot of y versus x.
Produces a stem plot of y versus x.
polar(theta,r,’type’)
semilogx(x,y)
semilogy(x,y)
stairs(x,y)
stem(x,y)
103
102
0.5x
y = 15x 0.37
y = 25e
x
102
101
y
y
y = 40(1.7)
0
1
x
2
3
101
100
10 –1
100
x
101
Figure 5.2–6 Two examples of exponential functions plotted with the semilogy function (the left-hand plot), and an example of a power function plotted with the loglog
function (right-hand) plot.
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Note that the two exponential functions y 25e 0.5x and y 40(1.7)x both produce
straight lines on a semilog plot with the y axis logarithmic. The power function
y 15x 0.37 produces a straight line on a log-log plot.
Stem, Stairs, and Bar Plots
MATLAB has several other plot types that are related to xy plots. These include
the stem, stairs, and bar plots. Their syntax is very simple, namely, stem(x,y),
stairs(x,y), and bar(x,y). See Table 5.2–3.
Separate y Axes
The plotyy function generates a graph with two y axes. The syntax plotyy
(x1,y1,x2,y2) plots y1 versus x1 with y axis labeling on the left, and plots y2
versus x2 with y axis labeling on the right. The syntax plotyy(x1,y1,x2,
y2,’type1’,’type2’) generates a ‘type1’ plot of y1 versus x1 with y axis
labeling on the left, and generates a ‘type2’ plot of y2 versus x2 with y axis labeling on the right. For example, plotyy(x1,y1,x2,y2,‘plot’,‘stem’)
uses plot(x1,y1) to generate a plot for the left axis, and stem(x2,y2) to generate a plot for the right axis. To see other variations of the plotyy function, type
help plotyy.
Polar Plots
Polar plots are two-dimensional plots made using polar coordinates. If the
polar coordinates are ( , r), where is the angular coordinate and r is the radial
coordinate of a point, then the command polar(theta,r) will produce the
polar plot. A grid is automatically overlaid on a polar plot. This grid consists of
concentric circles and radial lines every 30 . The title and gtext commands can be used to place a title and text. The variant command
polar(theta,r,’type’) can be used to specify the line type or data
marker, just as with the plot command.
Plotting Orbits
EXAMPLE 5.2–1
The equation
r =
p
1 -
cos
describes the polar coordinates of an orbit measured from one of the orbit’s two focal
points. For objects in orbit around the sun, the sun is at one of the focal points. Thus
r is the distance of the object from the sun. The parameters p and determine the size
of the orbit and its eccentricity, respectively. Obtain the polar plot that represents an
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Orbital Eccentricity = 0.5
90
4
120
60
3
2
150
30
1
180
0
210
330
240
300
270
Figure 5.2–7 A polar plot showing an orbit having an eccentricity of 0.5.
orbit having 0.5 and p 2 AU (AU stands for “astronomical unit”; 1 AU is the
mean distance from the sun to Earth). How far away does the orbiting object get from
the sun? How close does it approach Earth’s orbit?
■ Solution
Figure 5.2–7 shows the polar plot of the orbit. The plot was generated by the following
session.
>>theta = 0:pi/90:2*pi;
>>r = 2./(1-0.5*cos(theta));
>>polar(theta,r),title(‘Orbital Eccentricity = 0.5’)
The sun is at the origin, and the plot’s concentric circular grid enables us to determine that the closest and farthest distances the object is from the sun are approximately
1.3 and 4 AU. Earth’s orbit, which is nearly circular, is represented by the innermost
circle. Thus the closest the object gets to Earth’s orbit is approximately 0.3 AU. The
radial grid lines allow us to determine that when = 90 and 270 , the object is 2 AU
from the sun.
Error Bar Plots
Experimental data are often represented with plots containing error bars. The
bars show the estimated or calculated errors for each data point. They can also be
used to display the error in an approximate formula. For example, keeping two
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Approximation = 1 − x 2/2
2
cos(x)
1
0
−1
−2
−3
−4
0
0.5
1
1.5
x
2
2.5
3
3.5
Figure 5.2–8 Error bars for the approximation cos x ⬇ 1 x2兾2.
terms in the Taylor series expansion of cos x about x 0 gives cos x ⬇ 1 x 2兾2.
The following program creates the plot shown in Figure 5.2–8.
% errorbar example
x = linspace(0.1, pi, 20);
approx = 1 - x.^2/2;
error = approx - cos(x);
errorbar(x, cos(x), error), legend(‘cos(x)’), . . .
title(‘Approximation = 1 - x^2/2’)
There are more than 20 two-dimensional plot functions available in MATLAB.
We have shown the most important ones for engineering applications.
Test Your Understanding
T5.2–6 Plot the following functions using axes that will produce a straight-line plot.
The power function is y 2x0.5, and the exponential function is y 101x.
T5.2–7 Plot the function y 8x3 for 1 x 1 with a tick spacing of 0.25 on
the x axis and 2 on the y axis.
T5.2–8 The spiral of Archimedes is described by the polar coordinates ( , r),
where r a . Obtain a polar plot of this spiral for 0 4, with the
parameter a 2.
Publishing Reports Containing Graphics
Starting with MATLAB 7, the publish function is available for creating reports, which may have embedded graphics. Reports generated by the publish
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function may be exported to a variety of common formats including HTML
(Hyper Text Markup Language), which is used for Web-based reports; MS Word;
PowerPoint; and LATEX. To publish a report, do the following.
1. Open the Editor, type in the M- le that forms the basis of the report, and
save it. Use the double percent character (%%) to indicate a section heading in the report. This character marks the beginning of a new cell, which is
a group of commands. (Such a cell should not be confused with the cell
array data type covered in Section 2.6.) Enter any blank lines you wish to
appear in the report. Consider, as a very simple example, the following
sample le polyplot.m.
%% Example of Report Publishing:
% Plotting the cubic y = x^3 - 6x^2 + 10x+4.
%% Create the independent variable.
x = linspace(0, 4, 300); % Use 300 points between 0 and 4.
%% De ne the cubic from its coef cients.
p = [1, -6, 10, 4]; % p contains the coef cients.
%% Plot the cubic
plot(x, polyval(p, x)), xlabel(‘x’), ylabel(‘y’)
2. Run the le to check it for errors. (T o do this for a larger le, you may use
the cell mode of the Debugger to execute its each cell one at a time; see
Section 4.7.)
3. Use the publish and open functions to create the report in the desired format. Using our sample le, we can obtain a report in HTML format by typing
>>publish (‘polyplot’,’html’)
>>open html/polyplot.html
You should see a report like the one shown in Figure 5.2–9.
Instead of using the publish and open functions, you may select
Publish to HTML from the File menu in the Editor window. To publish to
another format, select instead Publish to and then choose the desired
format from the menu.
Once it is published in HTML, you may click on a section heading in the
Contents to go to that section. This is useful for larger reports.
If you want the equation to look professionally typeset, you may edit the
resulting report in the appropriate editor (say, MS Word or LATEX). For example, to set the cubic polynomial in the resulting LATEX le, use the commands
presented earlier in this section to replace the equation in the second line of the
report with
y = {\it x}^3 - 6{\it x}^2 + 10{\it x} + 4
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Example of Report Publishing:
Plotting the cubic y = xˆ3 6 xˆ2 + 10x + 4 .
Contents
• Create the independent variable.
• Define the cubic.
• Plot the cubic.
Create the independent variable.
x = linspace(0, 4, 300); % Use 300 points between 0 and 4.
Define the cubic.
p = [1, -6, 10, 4];
% p contains the coefficients.
Plot the cubic.
plot(x,polyval(p,x)),xlabel(’x’),ylabel(’y’)
12
11
10
9
y
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7
6
5
4
0
0.5
1
1.5
2
x
2.5
Figure 5.2–9 A sample report published from MATLAB.
3
3.5
4
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5.3 Interactive Plotting in MATLAB
The interactive plotting environment in MATLAB is a set of tools for
■
■
■
■
■
Creating different types of graphs,
Selecting variables to plot directly from the Workspace Browser,
Creating and editing subplots,
Adding annotations such as lines, arrows, text, rectangles, and ellipses, and
Editing properties of graphics objects, such as their color, line weight, and font.
The Plot Tools interface includes the following three panels associated with
a given gure.
■
■
■
The Figure Palette: Use this to create and arrange subplots, to view and
plot workspace variables, and to add annotations.
The Plot Browser: Use this to select and control the visibility of the axes
or graphics objects plotted in the gure, and to add data for plotting.
The Property Editor: Use this to set basic properties of the selected object
and to obtain access to all properties through the Property Inspector.
The Figure Window
When you create a plot, the Figure window appears with the Figure toolbar visible (see Figure 5.3–1). This window has eight menus.
The File Menu The File menu is used for saving and printing the gure. This
menu was discussed in Section 5.1 under Saving Figures and Exporting
Figures.
The Edit Menu You can use the Edit menu to cut, copy, and paste items, such
as legend or title text, that appear in the gure. Click on Figure Properties to
open the Property Editor—Figure dialog box to change certain properties of the
gure.
Three items on the Edit menu are very useful for editing the gure. Clicking
the Axes Properties item brings up the Property Editor—Axes dialog box.
Double-clicking on any axis also brings up this box. You can change the scale
type (linear, log, etc.), the labels, and the tick marks by selecting the tab for the
desired axis or the font to be edited.
The Current Object Properties item enables you to change the properties
of an object in the gure. To do this, rst click on the object, such as a plotted
Figure 5.3–1 The Figure toolbar displayed.
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line, then click on Current Object Properties in the Edit menu. You will see the
Property Editor—Lineseries dialog box that lets you change properties such as
line weight and color, data-marker type, and plot type.
Clicking on any text, such as that placed with the title, xlabel,
ylabel, legend, or gtext commands, and then selecting Current Object
Properties in the Edit menu bring up the Property Editor—Text dialog box,
which enables you to edit the text.
The View Menu The items on the View menu are the three toolbars (Figure
Toolbar, Plot Edit Toolbar, and Camera Toolbar), the Figure Palette, the Plot
Browser, and the Property Editor. These will be discussed later in this section.
The Insert Menu The Insert menu enables you to insert labels, legends, titles,
text, and drawing objects, rather than using the relevant commands from the
Command window. To insert a label on the y axis, for example, click on the Y Label
item on the menu; a box will appear on the y axis. Type the label in this box, and
then click outside the box to nish.
The Insert menu also enables you to insert arrows, lines, text, rectangles,
and ellipses in the gure. To insert an arrow, for example, click on the Arrow
item; the mouse cursor changes to a crosshair style. Then click the mouse button,
and move the cursor to create the arrow. The arrowhead will appear at the point
where you release the mouse button. Be sure to add arrows, lines, and other annotations only after you are nished moving or resizing your axes, because these
objects are not anchored to the axes. (They can be anchored to the plot by pinning; see the MATLAB Help.)
To delete or move a line or arrow, click on it, then press the Delete key to
delete it, or press the mouse button and move it to the desired location. The Axes
item lets you use the mouse to place a new set of axes within the existing plot.
Click on the new axes, and a box will surround them. Any further plot commands
issued from the Command window will direct the output to these axes.
The Light item applies to three-dimensional plots.
The Tools Menu The Tools menu includes items for adjusting the view (by
zooming and panning) and the alignment of objects on the plot. The Edit Plot
item starts the plot editing mode, which can also be started by clicking on the
northwest-facing arrow on the Figure toolbar. The Tools menu also gives access
to the Data Cursor, which is discussed later in this section. The last two items,
Basic Fitting and Data Statistics, will be discussed in Sections 6.3 and 7.1,
respectively.
Other Menus The Desktop menu enables you to dock the Figure window
within the desktop. The Window menu lets you switch between the Command
window and any other Figure windows. The Help menu accesses the general
MATLAB Help System as well as Help features speci c to plotting.
There are three toolbars available in the Figure window: the Figure toolbar,
the Plot Edit toolbar, and the Camera toolbar. The View menu lets you select
which ones you want to appear. We will discuss the Figure toolbar and the Plot
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Edit toolbar in this section. The Camera toolbar is useful for three-dimensional
plots, which are discussed at the end of this chapter.
The Figure Toolbar
To activate the Figure toolbar, select it from the View menu (see Figure 5.3–1).
The four leftmost buttons are for opening, saving, and printing the gure.
Clicking on the northwest-facing arrow button toggles the plot edit mode on
and off.
The Zoom-in and Zoom-out buttons let you obtain a close-up or faraway
view of the gure. The Pan and Rotate 3D buttons are used for three-dimensional
plots.
The Data Cursor button enables you to read data directly from a graph by
displaying the values of points you select on plotted lines, surfaces, images, and
so on.
The Insert Colorbar button inserts a color map strip in the graph and is useful for three-dimensional surface plots. The Insert Legend button enables you to
insert a legend in the plot. The last two buttons hide or show the plot tools and
dock the gure if it is undocked.
The Plot Edit Toolbar
Once a plot is in the window, you can display the Plot Edit toolbar from the View
menu. This toolbar is shown in Figure 5.3–2. You can enable plot editing by
clicking on the northwest-facing arrow on the Figure toolbar. Then double-click
on an axis, a plotted line, or a label to activate the appropriate property editor. To
add text that is not a label, title, or legend, click the button labeled T, move the
cursor to the desired location for the text, click the mouse button, and type the
text. When nished, click outside the text box and note that the nine leftmost buttons become highlighted and available. These enable you to modify the color,
font, and other attributes of the text.
To insert arrows, lines, rectangles, and ellipses, click on the appropriate button and follow the instructions given previously for the Insert menu.
The Plot Tools
Once a gure has been created, you can display any of or all three Plot Tools
(Figure Palette, Plot Browser, and Property Editor) by selecting them from
Figure 5.3–2 The Figure and Plot Edit toolbars displayed.
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Figure 5.3–3 The Figure window with the Plot Tools activated.
the View menu. You can also start the environment by rst creating a plot and
then clicking on the Show Plot Tools icon at the far left of the Figure toolbar
(see Figure 5.3–3), or by creating a gure with the plotting tools attached by
using the plottools command. Remove the tools by clicking on the Hide
Tools icon, which is second from the left.
Figure 5.3–3 shows the result of clicking on the plotted line after clicking the
Show Plot Tools icon. The plotting interface then displays the Property Editor—
Lineseries.
The Figure Palette
The Figure Palette contains three panels, which are selected and expanded by
clicking the appropriate button. Click on the grid icon in the New Subplots panel
to display the selector grid that enables you to specify the layout of the subplots.
In the Variables panel you can select a graphics function to plot the variable by
selecting the variable and right-clicking to display the context menu. This menu
contains a list of possible plot types based on the type of variable you select. You
can also drag the variable into an axes set, and MATLAB will select an appropriate plot type.
Selecting More Plots from the context menu activates the Plot Catalog
tool, which provides access to most of the plotting functions. After selecting a
plot category, and a plot type from that category, you will see its description in
the rightmost display. Type the name of one or more variables in the Plotted
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Variables eld, separated by commas, and they will be passed to the selected
plotting function as arguments. You can also type a MATLAB expression that
uses any workspace variables shown in the Figure Palette.
Click on the Annotations panel to display a menu of objects such as lines,
arrows, etc. Click on the desired object, and use the mouse to position and size it.
The Plot Browser
The Plot Browser provides a legend of all the graphs in the gure. For example, if
you plot an array with multiple rows and columns, the browser lists each axis and
the objects (lines, surfaces, etc.) used to create the graph. To set the properties of an
individual line, double-click on the line. Its properties are displayed in the Property
Editor—Lineseries box, which opens on the bottom of the gure.
If you select a line in the graph, the corresponding entry in the Plot Browser
is highlighted, indicating which column in the variable produced the line. The
check box next to each item in the browser controls the object’s visibility. For example, if you want to plot only certain columns of data, you can uncheck the
columns not wanted. The graph updates as you uncheck each box and rescales
the axes as required.
The Property Editor
The Property Editor enables you to access a subset of the selected object’s
properties. When no object is selected, the Property Editor displays the gure’ s
properties. There are several ways to display the Property Editor.
1. Double-click an object when plot edit mode is enabled.
2. Select an object and right-click to display its context menu, then select
Properties.
3. Select Property Editor from the View menu.
4. Use the propertyeditor command.
The Property Editor enables you to change the most commonly used object
properties. If you want to access all object properties, use the Property Inspector.
To display the Property Inspector, click the Inspector button on any Property
Editor panel. Use of this feature requires detailed knowledge of object properties
and handle graphics, and thus will not be covered here.
Recreating Graphs from M-Files
Once your graph is finished, you can generate MATLAB code to reproduce
the graph by selecting Generate M-File from the File menu. MATLAB creates a function that recreates the graph and opens the generated M-File in the
editor. This feature is particularly useful for capturing property settings and
other modifications made in the plot editor. You can also use the makemcode
function.
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Adding Data to Axes
The Plot Browser provides the mechanism by which you add data to axes. The
procedure is as follows:
1. Select a two-dimensional or three-dimensional axis from the New Subplots
subpanel.
2. After creating the axis, select it in the Plot Browser panel to enable the
Add Data button at the bottom of the panel.
3. Click the Add Data button to display the Add Data to Axes dialog box.
The Add Data to Axes dialog enables you to select a plot type and specify
the workspace variables to pass to the plotting function. You can also
specify a MATLAB expression, which is evaluated to produce the data
to plot.
5.4 Three-Dimensional Plots
MATLAB provides many functions for creating three-dimensional plots. Here
we will summarize the basic functions to create three types of plots: line plots,
surface plots, and contour plots. Information about the related functions is available in MATLAB Help (category graph3d).
Three-Dimensional Line Plots
Lines in three-dimensional space can be plotted with the plot3 function. Its
syntax is plot3(x,y,z). For example, the following equations generate a
three-dimensional curve as the parameter t is varied over some range:
x = e-0.05t sin t
y = e-0.05t cos t
z = t
If we let t vary from t 0 to t 10, the sine and cosine functions will vary
through ve cycles, while the absolute values of x and y become smaller as t
increases. This process results in the spiral curve shown in Figure 5.4–1, which
was produced with the following session.
>>t = 0:pi/50:10*pi;
>>plot3(exp(-0.05*t).*sin(t),exp(-0.05*t).*cos(t),t),...
xlabel(‘x’),ylabel(‘y’),zlabel(‘z’),grid
Note that the grid and label functions work with the plot3 function, and that
we can label the z axis by using the zlabel function, which we have seen for
the rst time. Similarly , we can use the other plot enhancement functions discussed in Sections 5.1 and 5.2 to add a title and text and to specify line type
and color.
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35
30
25
z
20
15
10
5
0
1
1
0.5
0.5
0
0
– 0.5
y
– 0.5
–1
–1
x
Figure 5.4–1 The curve x e0.05t sin t, y e0.05t cos t, z t plotted with the
plot3 function.
Surface Mesh Plots
The function z f (x, y) represents a surface when plotted on xyz axes, and the
mesh function provides the means to generate a surface plot. Before you can use
this function, you must generate a grid of points in the xy plane and then evaluate
the function f (x, y) at these points. The meshgrid function generates the grid.
Its syntax is [X,Y] = meshgrid(x,y). If x = xmin:xspacing:xmax
and y = ymin:yspacing:ymax, then this function will generate the coordinates of a rectangular grid with one corner at (xmin, ymin) and the opposite
corner at (xmax, ymax). Each rectangular panel in the grid will have a width
equal to xspacing and a depth equal to yspacing. The resulting matrices X and Y
contain the coordinate pairs of every point in the grid. These pairs are then used
to evaluate the function.
The function [X,Y] = meshgrid(x) is equivalent to [X,Y] =
meshgrid(x,x) and can be used if x and y have the same minimum values, the
same maximum values, and the same spacing. Using this form, you can type
[X,Y] = meshgrid(min:spacing:max), where min and max specify
the minimum and maximum values of both x and y and spacing is the desired
spacing of the x and y values.
After the grid is computed, you create the surface plot with the mesh function. Its syntax is mesh(x,y,z). The grid, label, and text functions can be used
with the mesh function. The following session shows how to generate the
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0.5
z
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0
– 0.5
2
2
1
1
0
0
–1
y
–1
–2
–2
Figure 5.4–2 A plot of the surface z xe
function.
[(xy2)2y2]
x
created with the mesh
surface plot of the function z xe[(xy2)2y2], for 2 x 2 and 2 y 2,
with a spacing of 0.1. This plot appears in Figure 5.4–2.
>>[X,Y] = meshgrid(-2:0.1:2);
>>Z = X.*exp(-((X-Y.^2).^2Y.^2));
>>mesh(X,Y,Z),xlabel(‘x’),ylabel(‘y’),zlabel(‘z’)
Be careful not to select too small a spacing for the x and y values for two reasons:
(1) Small spacing creates small grid panels, which make the surface dif cult to
visualize, and (2) the matrices X and Y can become too large.
The surf and surfc functions are similar to mesh and meshc except
that the former create a shaded surface plot. You can use the Camera toolbar
and some menu items in the Figure window to change the view and lighting of
the gure.
Contour Plots
Topographic plots show the contours of the land by means of constant elevation
lines. These lines are also called contour lines, and such a plot is called a contour
plot. If you walk along a contour line, you remain at the same elevation. Contour
plots can help you visualize the shape of a function. They can be created with the
contour function, whose syntax is contour(X,Y,Z). You use this function
the same way you use the mesh function; that is, rst use the meshgrid function
to generate the grid and then generate the function values. The following session
generates the contour plot of the function whose surface plot is shown in
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2
1.5
1
y
0.5
0
– 0.5
–1
– 1.5
–2
–2
–1.5
–1
–0.5
0
x
0.5
22
1
1.5
2
2
Figure 5.4–3 A contour plot of the surface z xe[(xy ) y ] created with the
contour function.
2 2
2
Figure 5.4–2, namely, z xe[(xy ) y ], for 2 x 2 and 2 y 2, with a
spacing of 0.1. This plot appears in Figure 5.4–3.
>>[X,Y] = meshgrid(-2:0.1:2);
>>Z = X.*exp(-((X-Y.^2).^2+Y.^2));
>>contour(X,Y,Z),xlabel(‘x’),ylabel(‘y’)
You can add labels to the contour lines. Type help clabel.
Contour plots and surface plots can be used together to clarify the function. For
example, unless the elevations are labeled on contour lines, you cannot tell whether
there is a minimum or a maximum point. However, a glance at the surface plot will
make this easy to determine. On the other hand, accurate measurements are not
possible on a surface plot; these can be done on the contour plot because no distortion is involved. Thus a useful function is meshc, which shows the contour lines
beneath the surface plot. The meshz function draws a series of vertical lines under
the surface plot, while the waterfall function draws mesh lines in one direction
only. The results of these functions are shown in Figure 5.4–4 for the function
z xe(x2y2).
Table 5.4–1 summarizes the functions introduced in this section. For other
three-dimensional plot types, type help specgraph.
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0.5
0.5
0
0
z
z
–0.5
2
2
0
y
–2 –2
–0.5
2
y
x
(a)
0.5
0
0
–0.5
2
2
0
–2
x
–0.5
2
2
0
0
–2
–2 –2
0
(b)
0.5
y
2
0
0
z
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x
(c)
–2 –2
0
x
(d)
Figure 5.4–4 Plots of the surface z xe(x y ) created with the mesh function and its
variant forms: meshc, meshz, and waterfall. (a) mesh, (b) meshc, (c) meshz,
(d) waterfall.
2
2
Table 5.4–1 Three-dimensional plotting functions
Function
Description
contour(x,y,z)
mesh(x,y,z)
meshc(x,y,z)
meshz(x,y,z)
Creates a contour plot.
Creates a three-dimensional mesh surface plot.
Same as mesh but draws a contour plot under the surface.
Same as mesh but draws a series of vertical reference
lines under the surface.
Creates a shaded three-dimensional mesh surface plot.
Same as surf but draws a contour plot under the surface.
Creates the matrices X and Y from the vectors x and y to
de ne a rectangular grid.
Same as [X,Y] = meshgrid(x,x).
Same as mesh but draws mesh lines in one direction only.
surf(x,y,z)
surfc(x,y,z)
[X,Y] = meshgrid(x,y)
[X,Y] = meshgrid(x)
waterfall(x,y,z)
Test Your Understanding
T5.4–1 Create a surface plot and a contour plot of the function z (x 2)2 2xy y2.
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5.5 Summary
This chapter explained how to use the powerful MATLAB commands to create
effective and pleasing two-dimensional and three-dimensional plots. The following guidelines will help you create plots that effectively convey the desired information.
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■
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■
■
■
Label each axis with the name of the quantity being plotted and its units!
Use regularly spaced tick marks at convenient intervals along each axis.
If you are plotting more than one curve or data set, label each on its plot or
use a legend to distinguish them.
If you are preparing multiple plots of a similar type or if the axes’ labels
cannot convey enough information, use a title.
If you are plotting measured data, plot each data point in a given set with
the same symbol, such as a circle, square, or cross.
If you are plotting points generated by evaluating a function (as opposed to
measured data), do not use a symbol to plot the points. Instead, connect the
points with solid lines.
Key Terms with Page References
Axis limits, 222
Contour plot, 248
Data symbol, 219
Overlay plot, 228
Polar plot, 236
Subplot, 226
Surface mesh plot, 247
Problems
You can nd the answers to problems marked with an asterisk at the end of the text.
Sections 5.1, 5.2, and 5.3
1.* Breakeven analysis determines the production volume at which the total
production cost is equal to the total revenue. At the breakeven point, there
is neither pro t nor loss. In general, production costs consist of xed costs
and variable costs. Fixed costs include salaries of those not directly involved with production, factory maintenance costs, insurance costs, and
so on. Variable costs depend on production volume and include material
costs, labor costs, and energy costs. In the following analysis, assume that
we produce only what we can sell; thus the production quantity equals the
sales. Let the production quantity be Q, in gallons per year.
Consider the following costs for a certain chemical product:
Fixed cost: $3 million per year.
Variable cost: 2.5 cents per gallon of product.
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The selling price is 5.5 cents per gallon.
Use these data to plot the total cost and the revenue versus Q, and
graphically determine the breakeven point. Fully label the plot and mark the
breakeven point. For what range of Q is production pro table? For what
value of Q is pro t a maximum?
2.
Consider the following costs for a certain chemical product:
Fixed cost: $2.045 million/year.
Variable costs:
Material cost: 62 cents per gallon of product.
Energy cost: 24 cents per gallon of product.
Labor cost: 16 cents per gallon of product.
Assume that we produce only what we sell. Let P be the selling price in
dollars per gallon. Suppose that the selling price and the sales quantity
Q are interrelated as follows: Q 6 106 1.1 106P. Accordingly,
if we raise the price, the product becomes less competitive and sales
drop.
Use this information to plot the xed and total variable costs versus
Q, and graphically determine the breakeven point(s). Fully label the plot
and mark the breakeven points. For what range of Q is production profitable? For what value of Q is pro t a maximum?
3.* a. Estimate the roots of the equation
x3 - 3x2 + 5x sin a
x
5
b + 3 = 0
4
4
by plotting the equation.
b. Use the estimates found in part a to nd the roots more accurately with
the fzero function.
4.
To compute the forces in structures, sometimes we must solve equations
similar to the following. Use the fplot function to nd all the positive
roots of this equation:
x tan x = 9
5.* Cables are used to suspend bridge decks and other structures. If a heavy
uniform cable hangs suspended from its two endpoints, it takes the shape
of a catenary curve whose equation is
x
y = a cosh a b
a
where a is the height of the lowest point on the chain above some horizontal reference line, x is the horizontal coordinate measured to the right from
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Problems
the lowest point, and y is the vertical coordinate measured up from the
reference line.
Let a 10 m. Plot the catenary curve for 20 x 30 m. How high
is each endpoint?
6.
Using estimates of rainfall, evaporation, and water consumption, the town
engineer developed the following model of the water volume in the
reservoir as a function of time
V(t) = 109 + 108(1 - e-t>100) - 107t
where V is the water volume in liters and t is time in days. Plot V(t)
versus t. Use the plot to estimate how many days it will take before the
water volume in the reservoir is 50 percent of its initial volume of 109 L.
7.
It is known that the following Leibniz series converges to the value /4 as
n→ .
n
1
2k + 1
k=0
Plot the difference between /4 and the sum S(n) versus n for 0 n 200.
S(n) = a ( - 1)k
8.
A certain shing vessel is initially located in a horizontal plane at x 0 and
y 10 mi. It moves on a path for 10 hr such that x t and y 0.5t2 10,
where t is in hours. An international shing boundary is described by the line
y 2x 6.
a. Plot and label the path of the vessel and the boundary.
b. The perpendicular distance of the point (x1, y1) from the line Ax By C 0 is given by
d =
Ax1 + By1 + C
1A2 + B2
where the sign is chosen to make d 0. Use this result to plot the distance
of the shing vessel from the shing boundary as a function of time for
0 t 10 hr.
9.
Plot columns 2 and 3 of the following matrix A versus column 1. The
data in column 1 are time (seconds). The data in columns 2 and 3 are
force (newtons).
0
5
A = £ 10
15
20
-7
-4
-1
1
2
6
3
9§
0
-1
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10.* Many applications use the following “small angle” approximation for the
sine to obtain a simpler model that is easy to understand and analyze. This
approximation states that sin x ⬇ x, where x must be in radians. Investigate the accuracy of this approximation by creating three plots. For the
rst, plot sin x and x versus x for 0 x 1. For the second, plot the
approximation error sin x x versus x for 0 x 1. For the third, plot
the relative error [sin(x) x]/sin(x) versus x for 0 x 1. How small
must x be for the approximation to be accurate within 5 percent?
11.
12.
13.
You can use trigonometric identities to simplify the equations that appear
in many applications. Con rm the identity tan(2 x) 2 tan x/(1 tan2 x)
by plotting both the left and the right sides versus x over the range
0 x 2.
The complex number identity eix cos x i sin x is often used to convert the solutions of equations into a form that is relatively easy to visualize. Con rm this identity by plotting the imaginary part versus the real
part for both the left and right sides over the range 0 x 2.
Use a plot over the range 0 x 5 to con rm that sin( ix) i sinh x.
14.* The function y(t) 1 ebt, where t is time and b > 0, describes many
processes, such as the height of liquid in a tank as it is being lled and
the temperature of an object being heated. Investigate the effect of the
parameter b on y (t). To do this, plot y versus t for several values of b on
the same plot. How long will it take for y(t) to reach 98 percent of its
steady-state value?
15.
The following functions describe the oscillations in electric circuits and
the vibrations of machines and structures. Plot these functions on the
same plot. Because they are similar, decide how best to plot and label
them to avoid confusion.
x(t) = 10e-0.5t sin(3t + 2)
y(t) = 7e-0.4t cos(5t - 3)
16.
In certain kinds of structural vibrations, a periodic force acting on
the structure will cause the vibration amplitude to repeatedly increase
and decrease with time. This phenomenon, called beating, also occurs
in musical sounds. A particular structure’s displacement is described by
y(t) =
1
f 12 - f 22
[cos( f2t) - cos( f1t)]
where y is the displacement in inches and t is the time in seconds. Plot
y versus t over the range 0 t 20 for f1 8 rad/sec and f2 1 rad/sec.
Be sure to choose enough points to obtain an accurate plot.
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Problems
17.* The height h(t) and horizontal distance x(t) traveled by a ball thrown at an
angle A with a speed ␷ are given by
h(t) = ␷t sin A -
1 2
gt
2
x(t) = ␷t cos A
18.
At Earth’s surface the acceleration due to gravity is g 9.81 m/s2.
a. Suppose the ball is thrown with a velocity ␷ 10 m/s at an angle of
35 . Use MATLAB to compute how high the ball will go, how far it
will go, and how long it will take to hit the ground.
b. Use the values of ␷ and A given in part a to plot the ball’s trajectory;
that is, plot h versus x for positive values of h.
c. Plot the trajectories for ␷ 10 m/s corresponding to ve values of the
angle A: 20 , 30 , 45 , 60 , and 70 .
d. Plot the trajectories for A 45 corresponding to ve values of the
initial velocity ␷: 10, 12, 14, 16, and 18 m/s.
The perfect gas law relates the pressure p, absolute temperature T, mass
m, and volume V of a gas. It states that
pV = mRT
The constant R is the gas constant. The value of R for air is 286.7
(N · m)/(kg · K). Suppose air is contained in a chamber at room temperature (20 C 293 K). Create a plot having three curves of the gas pressure
in N/m2 versus the container volume V in m3 for 20 V 100. The three
curves correspond to the following masses of air in the container: m 1 kg,
m 3 kg, and m 7 kg.
19.
Oscillations in mechanical structures and electric circuits can often be
described by the function
y(t) = e-t> sin( t + )
where t is time and is the oscillation frequency in radians per unit time.
The oscillations have a period of 2/ , and their amplitudes decay in time
at a rate determined by , which is called the time constant. The smaller is, the faster the oscillations die out.
a. Use these facts to develop a criterion for choosing the spacing of the
t values and the upper limit on t to obtain an accurate plot of y(t).
(Hint: Consider two cases: 4 2/ and 4 2/ .)
b. Apply your criterion, and plot y(t) for 10, , and 2.
c. Apply your criterion, and plot y(t) for 0.1, 8, and 2.
20.
When a constant voltage was applied to a certain motor initially at rest, its
rotational speed s(t) versus time was measured. The data appear in the
following table:
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Time (sec)
1
Speed (rpm) 1210
2
3
1866 2301
4
5
6
7
8
10
2564
2724
2881
2879
2915
3010
Determine whether the following function can describe the data. If so, nd
the values of the constants b and c.
s(t) = b(1 - ect)
21.
22.
The following table shows the average temperature for each year in a
certain city. Plot the data as a stem plot, a bar plot, and a stairs plot.
Year
2000
2001
2002
2003
2004
Temperature (⬚C)
21
18
19
20
17
A sum of $10 000 invested at 4 percent interest compounded annually will
grow according to the formula
y(k) = 104(1.04)k
where k is the number of years (k ⫽ 0, 1, 2, . . .). Plot the amount of
money in the account for a 10-year period. Do this problem with four
types of plots: the xy plot, the stem plot, the stairs plot, and the bar plot.
23.
The volume V and surface area A of a sphere of radius r are given by
V =
4 3
␲r
3
A = 4␲r 2
a. Plot V and A versus r in two subplots, for 0.1 ⱕ r ⱕ 100 m. Choose
axes that will result in straight-line graphs for both V and A.
b. Plot V and r versus A in two subplots, for 1 ⱕ A ⱕ 104 m2. Choose
axes that will result in straight-line graphs for both V and r.
24.
The current amount A of a principal P invested in a savings account
paying an annual interest rate r is given by
A = P a1 +
r nt
b
n
where n is the number of times per year the interest is compounded. For
continuous compounding, A ⫽ Pert. Suppose $10 000 is initially invested
at 3.5 percent (r ⫽ 0.035).
a. Plot A versus t for 0 ⱕ t ⱕ 20 years for four cases: continuous
compounding, annual compounding (n ⫽ 1), quarterly compounding
(n ⫽ 4), and monthly compounding (n ⫽ 12). Show all four cases on
the same subplot and label each curve. On a second subplot, plot the
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Problems
Source
+
i1
R1
v1
Load
v2
–
Figure P25
difference between the amount obtained from continuous compounding
and the other three cases.
b. Redo part a, but plot A versus t on log-log and semilog plots. Which
plot gives a straight line?
25.
Figure P25 is a representation of an electrical system with a power supply
and a load. The power supply produces the xed voltage ␷1 and supplies
the current i1 required by the load, whose voltage drop is ␷2. The current–
voltage relationship for a speci c load is found from experiments to be
i1 = 0.16(e0.12␷2 - 1)
Suppose that the supply resistance is R1 30 and the supply voltage is
␷1 15 V. To select or design an adequate power supply, we need to determine how much current will be drawn from the power supply when
this load is attached. Find the voltage drop ␷2 as well.
26.
The circuit shown in Figure P26 consists of a resistor and a capacitor and
is thus called an RC circuit. If we apply a sinusoidal voltage ␷i, called the
input voltage, to the circuit as shown, then eventually the output voltage
␷o will be sinusoidal also, with the same frequency but with a different
amplitude and shifted in time relative to the input voltage. Speci cally , if
␷i Ai sin t, then ␷o Ao sin( t ). The frequency–response plot is a
plot of Ao /Ai versus frequency . It is usually plotted on logarithmic axes.
+
R
C
vi
–
Figure P26
vo
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CHAPTER 5 Advanced Plotting
Upper-level engineering courses explain that for the RC circuit shown,
this ratio depends on and RC as follows:
Ao
1
= `
`
Ai
RCs + 1
where s i. For RC 0.1 s, obtain the log-log plot of Ao /Aiversus
and use it to nd the range of frequencies for which the output amplitude
Ao is less than 70 percent of the input amplitude Ai.
27.
An approximation to the function sin x is sin x ⬇ x x3Ⲑ6. Plot the sin x
function and 20 evenly spaced error bars representing the error in the
approximation.
Section 5.4
28.
The popular amusement ride known as the corkscrew has a helical shape.
The parametric equations for a circular helix are
x = a cos t
y = a sin t
z = bt
where a is the radius of the helical path and b is a constant that
determines the “tightness” of the path. In addition, if b 0, the helix has
the shape of a right-handed screw; if b 0, the helix is left-handed.
Obtain the three-dimensional plot of the helix for the following three
cases and compare their appearance with one another. Use 0 t 10
and a 1.
a. b 0.1
b. b 0.2
c. b 0.1
29.
A robot rotates about its base at 2 rpm while lowering its arm and extending
its hand. It lowers its arm at the rate of 120 per minute and extends its hand
at the rate of 5 m/min. The arm is 0.5 m long. The xyz coordinates of the
hand are given by
x = (0.5 + 5t) sin a
2
t b cos (4t)
3
y = (0.5 + 5t) sin a
2
t b sin (4t)
3
z = (0.5 + 5t) cos a
2
tb
3
where t is time in minutes.
Obtain the three-dimensional plot of the path of the hand for 0 t 0.2 min.
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Problems
30.
Obtain the surface and contour plots for the function z x2 2xy 4y2,
showing the minimum at x y 0.
31.
Obtain the surface and contour plots for the function z x2 2xy 3y2. This surface has the shape of a saddle. At its saddlepoint at x y 0,
the surface has zero slope, but this point does not correspond to either a
minimum or a maximum. What type of contour lines corresponds to a
saddlepoint?
32.
Obtain the surface and contour plots for the function z (x y2)(x 3y2).
This surface has a singular point at x y 0, where the surface has zero
slope, but this point does not correspond to either a minimum or a maximum. What type of contour lines corresponds to a singular point?
33.
A square metal plate is heated to 80 C at the corner corresponding to x y 1. The temperature distribution in the plate is described by
2
T = 80e-(x - 1) e-3(y - 1)
2
Obtain the surface and contour plots for the temperature. Label each axis.
What is the temperature at the corner corresponding to x y 0?
34.
The following function describes oscillations in some mechanical
structures and electric circuits.
z(t) = e-t> sin( t + )
In this function t is time, and is the oscillation frequency in radians per
unit time. The oscillations have a period of 2/ , and their amplitudes
decay in time at a rate determined by , which is called the time constant.
The smaller is, the faster the oscillations die out.
Suppose that 0, 2, and can have values in the range 0.5 10 sec. Then the preceding equation becomes
z(t) = e-t> sin(2t)
Obtain a surface plot and a contour plot of this function to help visualize
the effect of for 0 t 15 sec. Let the x variable be time t and the y
variable be .
35.
The following equation describes the temperature distribution in a at
rectangular metal plate. The temperature on three sides is held constant at
T1, and at T2 on the fourth side (see Figure P35). The temperature T (x, y) as
a function of the xy coordinates shown is given by
T(x, y) = (T2 - T1 )w(x,y) + T1
where
w(x,y) =
q
2
2
nx sinh(ny>L)
b
sin a
na
n
L
sinh(nW>L)
odd
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CHAPTER 5 Advanced Plotting
y
T2
W
T ( x ,y )
T1
T1
0
0
L
T1
x
Figure P35
The given data for this problem are T1 70 F, T2 200 F, and
W L 2 ft.
Using a spacing of 0.2 for both x and y, generate a surface mesh plot
and a contour plot of the temperature distribution.
36.
The electric potential eld V at a point, due to two charged particles, is
given by
V =
q1
q2
1
a
+ b
r
r2
4 0 1
where q1 and q2 are the charges of the particles in coulombs (C), r1 and r2
are the distances of the charges from the point (in meters), and 0 is the
permittivity of free space, whose value is
0 = 8.854 * 10
-12
C2>(N . m2)
Suppose the charges are q1 2 1010 C and q2 4 1010 C. Their
respective locations in the xy plane are (0.3, 0) and (0.3, 0) m. Plot the
electric potential eld on a three-dimensional surface plot with V plotted on
the z axis over the ranges 0.25 x 0.25 and 0.25 y 0.25. Create
the plot in two ways: (a) by using the surf function and (b) by using the
meshc function.
37.
Refer to Problem 25 of Chapter 4. Use the function le created for that
problem to generate a surface mesh plot and a contour plot of x versus h
and W for 0 W 500 N and for 0 h 2 m. Use the values k1 104 N/m, k2 1.5 104 N/m, and d 0.1 m.
38.
Refer to Problem 28 of Chapter 4. To see how sensitive the cost is to
location of the distribution center, obtain a surface plot and a contour plot
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Problems
of the total cost as a function of the x and y coordinates of the distribution
center location. How much would the cost increase if we located the
center 1 mi in any direction from the optimal location?
39.
Refer to Example 3.2–1. Use a surface plot and a contour plot of the perimeter length L as a function of d and over the ranges 1 d 30 ft and 0.1 1.5 rad. Are there valleys other than the one corresponding to d 7.5984 and 1.0472? Are there any saddle points?
261
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Copyright Royalty-Free/CORBIS
Engineering in the
21st Century. . .
Virtual Prototyping
T
o many people, computer-aided design (CAD) or computer-aided engineering (CAE) means creating engineering drawings. However, it means
much more. Engineers can use computers to determine the forces, voltages, currents, and so on that might occur in a proposed design. Then they can
use this information to make sure the hardware can withstand the predicted
forces or supply the required voltages or currents. Engineers are just beginning
to use the full potential of CAE.
The normal stages in the development of a new vehicle, such as an aircraft,
formerly consisted of aerodynamic testing a scale model; building a full-size
wooden mock-up to check for pipe, cable, and structural interferences; and
nally building and testing a prototype, the rst complete vehicle. CAE is changing the traditional development cycle. The Boeing 777 is the rst aircraft to be
designed and built using CAE, without the extra time and expense of building a
mock-up. The design teams responsible for the various subsystems, such as aerodynamics, structures, hydraulics, and electrical systems, all had access to the
same computer database that described the aircraft. Thus when one team made a
design change, the database was updated, allowing the other teams to see
whether the change affected their subsystem. This process of designing and testing with a computer model has been called virtual prototyping. Virtual prototyping reduced engineering changes, errors, and rework by 50 percent and greatly
enhanced the manufacturability of the airplane. When production began, the
parts went together easily.
MATLAB is a powerful tool for many CAE applications. It complements
geometric modeling packages because it can do advanced calculations that such
packages cannot do. ■
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C H A P T E R
6
Model Building and
Regression
OUTLINE
6.1 Function Discovery
6.2 Regression
6.3 The Basic Fitting Interface
6.4 Summary
Problems
An important application of the plotting techniques covered in Chapter 5 is
function discovery, the technique for using data plots to obtain a mathematical
function or “mathematical model” that describes the process that generated the
data. This is the topic of Section 6.1. A systematic way of nding an equation that
best ts the data is regression (also called the least-squares method). Regression
is treated in Section 6.2. Section 6.3 introduces the MATLAB Basic Fitting
interface, which supports regression.
6.1 Function Discovery
Function discovery is the process of nding, or “discovering,” a function that can
describe a particular set of data. The following three function types can often describe physical phenomena.
1. The linear function: y(x) mx b. Note that y(0) b.
2. The power function: y(x) bxm. Note that y(0) 0 if m 0, and y(0) if m 0.
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CHAPTER 6 Model Building and Regression
3. The exponential function: y(x) b(10)mx or its equivalent form y bemx, where e is the base of the natural logarithm (ln e 1). Note that
y(0) b for both forms.
Each function gives a straight line when plotted using a speci c set of axes:
1. The linear function y mx b gives a straight line when plotted on
rectilinear axes. Its slope is m and its intercept is b.
2. The power function y bxm gives a straight line when plotted on log-log
axes.
3. The exponential function y b(10)mx and its equivalent form y bemx give
a straight line when plotted on a semilog plot whose y axis is logarithmic.
We look for a straight line on the plot because it is relatively easy to recognize,
and therefore we can easily tell whether the function will t the data well.
Use the following procedure to nd a function that describes a given set of
data. We assume that one of the function types (linear, exponential, or power) can
describe the data.
1. Examine the data near the origin. The exponential function can never
pass through the origin (unless of course b 0, which is a trivial case).
(See Figure 6.1–1 for examples with b 1.) The linear function can pass
through the origin only if b 0. The power function can pass through the
origin but only if m 0. (See Figure 6.1–2 for examples with b 1.)
The Exponential Function y = 10mx
4
m=2
3.5
m=1
3
2.5
y
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2
1.5
m=0
1
0.5
m = –1
m = –2
0
0
0.1
0.2
0.3
0.4
0.5
x
0.6
Figure 6.1–1 Examples of exponential functions.
0.7
0.8
0.9
1
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6.1
Function Discovery
The Power Function y = x m
4
3.5
m=2
m=1
3
y
2.5
m = 0.5
2
1.5
m=0
1
0.5
0
0
m = – 0.5
0.5
1
1.5
2
x
2.5
3
3.5
4
Figure 6.1–2 Examples of power functions.
2. Plot the data using rectilinear scales. If the data form a straight line, then
the data can be represented by the linear function and you are nished.
Otherwise, if you have data at x 0, then
a. If y(0) = 0, try the power function.
b. If y(0) Z 0, try the exponential function.
If data are not given for x 0, proceed to step 3.
3. If you suspect a power function, plot the data using log-log scales. Only a
power function will form a straight line on a log-log plot. If you suspect an
exponential function, plot the data using the semilog scales. Only an exponential function will form a straight line on a semilog plot.
4. In function discovery applications, we use the log-log and semilog plots
only to identify the function type, but not to nd the coef cients b and m.
The reason is that it is dif cult to interpolate on log scales.
We can nd the values of b and m with the MATLAB poly t function.
This function nds the coef cients of a polynomial of speci ed degree n that
best ts the data, in the so-called least-squares sense. The syntax appears in
Table 6.1–1. The mathematical foundation of the least-squares method is presented in Section 6.2.
Because we are assuming that our data will form a straight line on a
rectilinear, semilog, or log-log plot, we are interested only in a polynomial
that corresponds to a straight line, that is, a rst-degree polynomial, which we
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Table 6.1–1 The poly t function
Command
Description
p = poly t(x,y,n)
Fits a polynomial of degree n to data described by the
vectors x and y, where x is the independent variable.
Returns a row vector p of length n 1 that contains the
polynomial coef cients in order of descending powers.
will denote as w p1z p2. Thus, referring to Table 6.1–1, we see that the
vector p will be [p1, p2] if n is 1. This polynomial has a different interpretation in each of the three cases:
■
The linear function: y mx b. In this case the variables w and z in the
polynomial w p1z p2 are the original data variables x and y, and we can
nd the linear function that ts the data by typing p = poly t(x,y,1).
The rst element p1 of the vector p will be m, and the second element p2 will
be b.
■
The power function: y bxm. In this case log10 y m log10 x log10 b,
which has the form w p1z p2, where the polynomial variables w and z
are related to the original data variables x and y by w log10 y and z log10 x. Thus we can nd the power function that ts the data by typing p =
poly t(log10(x), log10(y),1). The rst element p1 of the vector
p will be
m, and the second element p2 will be log10 b. We can nd b from
p
b = 10 2.
■
The exponential function: y b(10)mx. In this case log10 y mx log10 b,
which has the form w p1z p2, where the polynomial variables w and z
are related to the original data variables x and y by w log10 y and z x.
Thus we can nd the exponential function that ts the data by typing p =
poly t(x, log10(y),1). The rst element p1 of the vector p willpbe
m, and the second element p2 will be log10 b. We can nd b from b = 10 2.
Temperature Dynamics
EXAMPLE 6.1–1
The temperature of coffee cooling in a porcelain mug at room temperature (68F) was
measured at various times. The data follow.
Time t (sec)
Temperature T (ⴗF)
0
620
2266
3482
145
130
103
90
Develop a model of the coffee’s temperature as a function of time, and use the model to
estimate how long it took the temperature to reach 120F.
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6.1
2
10
Relative Temperature (deg F)
Relative Temperature (deg F)
80
70
60
50
40
30
20
Function Discovery
1
0
1000
2000
Time (sec)
3000
10
4000
2
1000
2000
Time (sec)
3000
4000
0
1000
2000
Time (sec)
3000
4000
150
Relative Temperature (deg F)
10
0
Temperature (deg F)
140
130
120
110
100
90
1
10
0
1000
2000
Time (sec)
3000
4000
80
Figure 6.1–3 Temperature of a cooling cup of coffee, plotted on various coordinates.
■ Solution
Because T(0) is nite but nonzero, the power function cannot describe these data, so we do
not bother to plot the data on log-log axes. Common sense tells us that the coffee will cool
and its temperature will eventually equal the room temperature. So we subtract the room
temperature from the data and plot the relative temperature, T 68, versus time. If the relative temperature is a linear function of time, the model is T 68 mt b. If the relative
temperature is an exponential function of time, the model is T
68 b(10)mt. Figure
6.1–3 shows the plots used to solve the problem. The following MATLAB script le generates the top two plots. The time data are entered in the array time, and the temperature
data are entered in temp.
% Enter the data.
time = [0,620,2266,3482];
temp = [145,130,103,90];
% Subtract the room temperature.
temp = temp - 68;
% Plot the data on rectilinear scales.
subplot(2,2,1)
plot(time,temp,time,temp,’o’),xlabel(‘Time (sec)’),.. .
ylabel(‘Relative Temperature (deg F)’)
%
% Plot the data on semilog scales.
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CHAPTER 6 Model Building and Regression
268
subplot(2,2,2)
semilogy(time,temp,time,temp,’o’),xlabel(‘Time (sec)’),. . .
ylabel(‘Relative Temperature (deg F)’)
The data form a straight line on the semilog plot only (the top right plot). Thus the data
can be described with the exponential function T 68 b(10)mt. Using the poly t
command, the following lines can be added to the script le.
% Fit a straight line to the transformed data.
p = poly t(time,log10(temp),1);
m = p(1)
b = 10^p(2)
The computed values are m 1.5557
10 4 and b 77.4469. Thus our derived
mt
model is T 68 b(10) . To estimate how long it will take for the coffee to cool to
120F, we must solve the equation 120 68 b(10)mt for t. The solution is t [log10 (120 68) log10 (b)]/m. The MATLAB command for this calculation is shown in
the following script le, which is a continuation of the previous script and produces the
bottom two subplots shown in Figure 6.1–3.
% Compute the time to reach 120 degrees.
t_120 = (log10(120-68)-log10(b))/m
% Show the derived curve and estimated point on semilog scales.
t = 0:10:4000;
T = 68+b*10.^(m*t);
subplot(2,2,3)
semilogy(t,T-68,time,temp,’o’,t_120,120-68,’+’),
xlabel(‘Time (sec)’),...
ylabel(‘Relative Temperature (deg F)’)
%
% Show the derived curve and estimated point on rectilinear scales.
subplot(2,2,4)
plot(t,T,time,temp+68,’o’,t_120,120,’+’),xlabel(‘Time (sec)’),...
ylabel(‘Temperature (deg F)’)
The computed value of t_120 is 1112. Thus the time to reach 120 F is 1112 sec. The
plot of the model, along with the data and the estimated point (1112, 120) marked with
a sign, is shown in the bottom two subplots in Figure 6.1–3. Because the graph of
our model lies near the data points, we can treat its prediction of 1112 sec with some
con dence.
EXAMPLE 6.1–2
Hydraulic Resistance
A 15-cup coffee pot (see Figure 6.1–4) was placed under a water faucet and lled to the
15-cup line. With the outlet valve open, the faucet’s ow rate was adjusted until the water
level remained constant at 15 cups, and the time for one cup to ow out of the pot was
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6.1
Function Discovery
Figure 6.1–4 An experiment to verify
Torricelli’s principle.
measured. This experiment was repeated with the pot lled to the various levels shown in
the following table:
Liquid volume V (cups)
Time to ll 1 cup t (sec)
15
12
9
6
6
7
8
9
(a) Use the preceding data to obtain a relation between the ow rate and the number
of cups in the pot. (b) The manufacturer wants to make a 36-cup pot using the same
outlet valve but is concerned that a cup will ll too quickly , causing spills. Extrapolate the
relation developed in part (a) and predict how long it will take to ll one cup when the pot
contains 36 cups.
■ Solution
(a) Torricelli’s principle in hydraulics states that f rV1/2, where f is the ow rate through
the outlet valve in cups per second, V is the volume of liquid in the pot in cups, and r is a
constant whose value is to be found. We see that this relation is a power function where
the exponent is 0.5. Thus if we plot log10 ( f ) versus log10 (V ), we should obtain a straight
line. The values for f are obtained from the reciprocals of the given data for t. That is,
f 1/t cups per second.
The MATLAB script le follows. The resulting plots appear in Figure 6.1–5. The
volume data are entered in the array cups, and the time data are entered in
meas_times.
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Flow Rate (cups/sec)
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10
1
Volume (cups)
10
10
Fill Time per Cup (sec)
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8
6
4
2
0
5
10
15
20
25
Volume (cups)
30
35
Figure 6.1–5 Flow rate and ll time for a cof fee pot.
% Data for the problem.
cups = [6,9,12,15];
meas_times = [9,8,7,6];
meas_ ow = 1./meas_times;
%
% Fit a straight line to the transformed data.
p = poly t(log10(cups),log10(meas_ ow),1);
coeffs = [p(1),10^p(2)];
m = coeffs(1)
b = coeffs(2)
%
% Plot the data and the tted line on a loglog plot to see
% how well the line ts the data.
x = 6:0.01:40;
y = b*x.^m;
subplot(2,1,1)
loglog(x,y,cups,meas_ ow,’o’),grid,xlabel(‘Volume (cups)’),.. .
ylabel(‘Flow Rate (cups/sec)’),axis([5 15 0.1 0.3])
The computed values are m 0.433 and b 0.0499, and our derived relation is f 0.0499V 0.433. Because the exponent is 0.433, not 0.5, our model does not agree exactly
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with Torricelli’s principle, but it is close. Note that the rst plot in Figure 6.1–5 shows
that the data points do not lie exactly on the tted straight line. In this application it is
dif cult to measure the time to ll one cup with an accuracy greater than an integer
second, so this inaccuracy could have caused our result to disagree with that predicted
by Torricelli.
(b) Note that the ll time is 1 /f, the reciprocal of the ow rate. The remainder of the
MATLAB script uses the derived ow rate relation f 0.0499V 0.433 to plot the extrapolated ll-time curve 1/ f versus t.
% Plot the ll time curve extrapolated to 36 cups.
subplot(2,1,2)
plot(x,1./y,cups,meas_times,’o’),grid,xlabel(‘Volume(cups)’),...
ylabel(‘Fill Time per Cup (sec)’),axis([5 36 0 10])
%
% Compute the ll time for V = 36 cups.
ll_time = 1/(b*36^m)
The predicted ll time for 1 cup is 4.2 sec. The manufacturer must now decide if
this time is suf cient for the user to avoid over lling. (In fact, the manufacturer did
construct a 36-cup pot, and the ll time is approximately 4 sec, which agrees with our
prediction.)
6.2 Regression
In Section 6.1 we used the MATLAB function poly t to perform regression
analysis with functions that are linear or could be converted to linear form by a
logarithmic or other transformation. The poly t function is based on the leastsquares method, which is also called regression. We now show how to use this
function to develop polynomial and other types of functions.
The Least-Squares Method
Suppose we have the three data points given in the following table, and we need
to determine the coef cients of the straight line y mx b that best t the following data in the least-squares sense.
x
y
0
5
10
2
6
11
According to the least-squares criterion, the line that gives the best t is the
one that minimizes J, the sum of the squares of the vertical differences between
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CHAPTER 6 Model Building and Regression
RESIDUALS
the line and the data points. These differences are called the residuals. Here there
are three data points, and J is given by
3
J = a (mxi + b - yi)2
i=1
= (0m + b - 2)2 + (5m + b - 6)2 + (10m + b - 11)2
The values of m and b that minimize J are found by setting the partial derivatives J/ m and J/ b equal to zero.
J
= 250m + 30b - 280 = 0
m
J
= 30m + 6b - 38 = 0
b
These conditions give two equations that must be solved for the two unknowns m
and b. The solution is m 0.9 and b 11/6. The best straight line in the leastsquares sense is y 0.9x 11/6. If we evaluate this equation at the data values x 0, 5, and 10, we obtain the values y 1.833, 6.333, 10.8333. These values are different from the given data values y 2, 6, 11 because the line is not a perfect t to
the data. The value of J is J (1.833 2)2 (6.333 6)2 (10.8333 11)2 0.16656689. No other straight line will give a lower value of J for these data.
In general, for the polynomial a1xn a2xn 1 · · · an x an1, the sum
of the squares of the residuals for m data points is
m
J = a (a1x n + a2x n - 1 + Á + anx + an + 1 - yi)2
i=1
The values of the n 1 coef cients ai that minimize J can be found by solving a set
of n 1 linear equations. The poly t function provides this solution. Its syntax
is p = poly t(x,y,n). Table 6.2–1 summarizes the poly t and polyval
functions.
Consider the data set where x 1, 2, 3, . . . , 9 and y 5, 6, 10, 20, 28, 33,
34, 36, 42. The following script le computes the coef cients of the rst- through
fourth-degree polynomials for these data and evaluates J for each polynomial.
x = 1:9;
y = [5,6,10,20,28,33,34,36,42];
for k = 1:4
coeff = poly t(x,y,k)
J(k) = sum((polyval(coeff,x)-y).^2)
end
The J values are, to two signi cant gures, 72, 57, 42, and 4.7. Thus the value
of J decreases as the polynomial degree is increased, as we would expect. Figure 6.2–1 shows this data and the four polynomials. Note now the t improves
with the higher-degree polynomial.
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273
Table 6.2–1 Functions for polynomial regression
Command
Description
p = poly t(x,y,n)
Fits a polynomial of degree n to data described by the vectors x and
y, where x is the independent variable. Returns a row vector p of
length n+1 that contains the polynomial coef cients in order of
descending powers.
Fits a polynomial of degree n to data described by the vectors x and
y, where x is the independent variable. Returns a row vector p of
length n+1 that contains the polynomial coef cients in order of
descending powers and a structure s for use with polyval to
obtain error estimates for predictions. The optional output variable
mu is a two-element vector containing the mean and standard
deviation of x.
Uses the optional output structure s generated by [p,s,mu] =
poly t(x,y,n) to generate error estimates. If the errors in the
data used with poly t are independent and normally distributed
with constant variance, at least 50 percent of the data will lie within
the band y delta.
[p,s, mu] = poly t(x,y,n)
[y,delta] = polyval(p,x,s,mu)
First Degree
Second Degree
40
40
30
30
y
50
y
50
20
20
10
10
0
0
5
x
0
0
10
Third Degree
5
x
10
Fourth Degree
40
40
30
30
y
50
y
50
20
20
10
10
0
0
5
x
10
0
0
5
x
Figure 6.2–1 Regression using polynomials of rst through fourth degree.
10
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80
60
40
y
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0
–20
–40
0
0.5
1
1.5
2
2.5
x
3
3.5
4
4.5
5
Figure 6.2–2 An example of a fth-degree polynomial that passes through all six data
points but exhibits large excursions between points.
Caution: It is tempting to use a high-degree polynomial to obtain the best
possible t. However , there are two dangers in using high-degree polynomials.
High-degree polynomials often exhibit large excursions between the data
points and thus should be avoided if possible. Figure 6.2–2 shows an example
of this phenomenon. The second danger with using high-degree polynomials is
that they can produce large errors if their coef cients are not represented with
a large number of signi cant gures. In some cases it might not be possible to
t the data with a low-degree polynomial. In such cases we might be able to
use several cubic polynomials. This method, called cubic splines, is covered in
Chapter 7.
Test Your Understanding
T6.2–1 Obtain and plot the rst- through fourth-degree polynomials for the following data: x 0, 1, . . . , 5 and y 0, 1, 60, 40, 41, and 47. Find the
coef cients and the J values.
(Answer: The polynomials are 9.5714x 7.5714; 3.6964x2 28.0536x 4.7500; 0.3241x3 6.1270x2 32.4934x 5.7222; and
2.5208x4 24.8843x3 71.2986x2 39.5304x 1.4008. The corresponding J values are 1534, 1024, 1017, and 495, respectively.)
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Fitting Other Functions
Given the data (y, z), the logarithmic function y m ln z b can be converted to
a rst-degree polynomial by transforming the z values into x values by the transformation x ln z. The resulting function is y mx b.
Given the data (y, z), the function y b(10) m/z can be converted to an exponential function by transforming the z values by the transformation x 1/z.
Given the data (␷, x), the function ␷ 1/(mx b) can be converted to a
rst- degree polynomial by transforming the ␷ data values with the transformation
y 1/␷. The resulting function is y mx b.
To see how to obtain a function y kx that passes through the origin, see
Problem 8.
The Quality of a Curve Fit
The least-squares criterion used to t a function f (x) is the sum of the squares of
the residuals J. It is de ned as
m
J = a [ f(xi) - yi ]2
(6.2–1)
i=1
We can use the J value to compare the quality of the curve t for two or more
functions used to describe the same data. The function that gives the smallest
J value gives the best t.
We denote the sum of the squares of the deviation of the y values from their
mean y by S, which can be computed from
m
S = a (yi - y)2
(6.2–2)
i=1
This formula can be used to compute another measure of the quality of the curve
t, the coef cient of determination, also known as the r-squared value. It is
de ned as
r2 = 1 -
J
S
(6.2–3)
For a perfect t, J 0 and thus r2 1. Thus the closer r2 is to 1, the better
the t. The largest r2 can be is 1. The value of S indicates how much the data is
spread around the mean, and the value of J indicates how much of the data
spread is unaccounted for by the model. Thus the ratio J / S indicates the fractional variation unaccounted for by the model. It is possible for J to be larger
than S, and thus it is possible for r2 to be negative. Such cases, however, are indicative of a very poor model that should not be used. As a rule of thumb, a
good t accounts for at least 99 percent of the data variation. This value corresponds to r2 0.99.
For example, the following table gives the values of J, S, and r2 for the
rst- through fourth-degree polynomials used to t the data x 1, 2, 3, . . . , 9
and y 5, 6, 10, 20, 28, 33, 34, 36, 42.
COEFFICIENT OF
DETERMINATION
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Degree n
J
S
r2
1
2
3
4
72
57
42
4.7
1562
1562
1562
1562
0.9542
0.9637
0.9732
0.9970
Because the fourth-degree polynomial has the largest r2 value, it represents the
data better than the representation from rst- through third-degree polynomials,
according to the r2 criterion.
To calculate the values of S and r2, add the following lines to the end of the
script le shown on page 272.
mu = mean(y);
for k=1:4
S(k) = sum((y-mu).^2);
r2(k) = 1 - J(k)/S(k);
end
S
r2
Scaling the Data
The effect of computational errors in computing the coef cients can be lessened
by properly scaling the x values. When the function poly t(x,y,n) is executed, it will issue a warning message if the polynomial degree n is greater than
or equal to the number of data points (because there will not be enough equations
for MATLAB to solve for the coef cients), or if the vector x has repeated, or
nearly repeated, points, or if the vector x needs centering and/or scaling. The
alternate syntax [p, s, mu] = poly t(x,y,n) nds the coef cients p
of a polynomial of degree n in terms of the variable
xN = (x -
x)>x
The output variable mu is a two-element vector [ x, x], where x is the mean
of x and x is the standard deviation of x (the standard deviation is discussed
in Chapter 7).
You can scale the data yourself before using poly t. Some common scaling methods are
xN = x - xmin
or
xN = x -
or
xN =
if the range of x is small, or
xN =
if the range of x is large.
x
xmax
x
xmean
x
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Estimation of Traf c Flow
277
EXAMPLE 6.2–1
The following data give the number of vehicles (in millions) crossing a bridge each year
for 10 years. Fit a cubic polynomial to the data and use the t to estimate the ow in the
year 2010.
Year
2000
2001
2002
2003
2004
2005
2006
2007
2008
2009
Vehicle ow
(millions)
2.1
3.4
4.5
5.3
6.2
6.6
6.8
7
7.4
7.8
■ Solution
If we attempt to t a cubic to these data, as in the following session, we get a warning
message.
>>Year = 2000:2009;
>>Veh_Flow = [2.1,3.4,4.5,5.3,6.2,6.6,6.8,7,7.4,7.8];
>>p = poly t(Year,Veh_Flow,3)
Warning: Polynomial is badly conditioned.
The problem is caused by the large values of the independent variable Year. Because
their range is small, we can simply subtract 2000 from each value. Continue the session
as follows.
>>x = Year-2000; y = Veh_Flow;
>>p = poly t(x,y,3)
p =
0.0087
-0.1851
1.5991 2.0362
>>J = sum((polyval(p,x)-y).^2);
>>S = sum((y-mean(y)).^2);
>>r2 = 1 - J/S
r2 =
0.9972
Thus the polynomial t is good because the coef cient of determination is 0.9972. The
corresponding polynomial is
f = 0.0087(t - 2000)3 - 0.1851(t - 2000)2 + 1.5991(t - 2000) + 2.0362
where f is the traf c ow in millions of vehicles and t is the time in years measured from 0.
We can use this equation to estimate the ow at the year 2010 by substituting t 2010,
or by typing in MATLAB polyval(p,10). Rounded to one decimal place, the answer is 8.2 million vehicles.
Using Residuals
We now show how to use the residuals as a guide to choosing an appropriate function to describe the data. In general, if you see a pattern in the plot of the residuals, it indicates that another function can be found to describe the data better.
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Modeling Bacteria Growth
EXAMPLE 6.2–2
The following table gives data on the growth of a certain bacteria population with time.
Fit an equation to these data.
Time (min)
Bacteria (ppm)
Time (min)
Bacteria (ppm)
0
1
2
3
4
5
6
7
8
9
6
13
23
33
54
83
118
156
210
282
10
11
12
13
14
15
16
17
18
19
350
440
557
685
815
990
1170
1350
1575
1830
■ Solution
We try three polynomial ts (linear , quadratic, and cubic) and an exponential t. The script
le is given below . Note that we can write the exponential form as y b(10)mt 10mta,
where b 10a.
% Time data
x = 0:19;
% Population data
y = [6,13,23,33,54,83,118,156,210,282,...
350,440,557,685,815,990,1170,1350,1575,1830];
% Linear t
p1 = poly t(x,y,1);
% Quadratic t
p2 = poly t(x,y,2);
% Cubic t
p3 = poly t(x,y,3);
% Exponential t
p4 = poly t(x,log10(y),1);
% Residuals
res1 = polyval(p1,x)-y;
res2 = polyval(p2,x)-y;
res3 = polyval(p3,x)-y;
res4 = 10.^polyval(p4,x)-y;
You can then plot the residuals as shown in Figure 6.2–3. Note that there is a de nite
pattern in the residuals of the linear t. This indicates that the linear function cannot
match the curvature of the data. The residuals of the quadratic t are much smaller , but
there is still a pattern, with a random component. This indicates that the quadratic
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400
60
200
40
Residuals (ppm)
Residuals (ppm)
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Linear
0
–200
–400
–600
5
10
t (min)
15
10
Residuals (ppm)
Residuals (ppm)
20
15
20
Quadratic
0
5
10
t (min)
1500
5
0
–5
Cubic
–10
–15
15
0
–20
–60
20
Regression
20
–40
0
6.2
0
5
10
t (min)
15
20
1000
Exponential
500
0
–500
0
5
10
t (min)
Figure 6.2–3 Residual plots for the four models.
function also cannot match the curvature of the data. The residuals of the cubic t are
even smaller, with no strong pattern and a large random component. This indicates that
a polynomial degree higher than 3 will not be able to match the data curvature any better than the cubic. The residuals for the exponential are the largest of all, and indicate a
poor t. Note also how the residuals systematically increase with t, indicating that the
exponential cannot describe the data’s behavior after a certain time.
Thus the cubic is the best t of the four models considered. Its coef cient of determination is r2 0.9999. The model is
y = 0.1916t3 + 1.2082t2 + 3.607t + 7.7307
where y is the bacteria population in ppm and t is time in minutes.
Multiple Linear Regression
Suppose that y is a linear function of two or more variables x1, x2, . . . , for
example, y a0 a1x1 a2x2. To find the coefficient values a0, a1, and a2 to
fit a set of data (y, x1, x2) in the least-squares sense, we can make use of the
fact that the left-division method for solving linear equations uses the leastsquares method when the equation set is overdetermined. To use this method,
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let n be the number of data points and write the linear equation in matrix form as
Xa y, where
a0
a = a1
a2
1
1
X = 1
Á
x11
x12
x13
Á
x21
x22
x23
Á
1
x1n
x2n
y =
y1
y2
y3
Á
yn
where x1i, x2i, and yi are the data, i 1, . . . , n. The solution for the coef cients
is given by a = X\y.
Breaking Strength and Alloy Composition
EXAMPLE 6.2–3
We want to predict the strength of metal parts as a function of their alloy composition.
The tension force y required to break a steel bar is a function of the percentage x1 and x2
of each of two alloying elements present in the metal. The following table gives some
pertinent data. Obtain a linear model y a0 a1x1 a2x2 to describe the relationship.
Breaking strength (kN)
y
% of element 1
x1
%of element 2
x2
0
1
2
3
5
7
8
11
7.1
19.2
31
45
■ Solution
The script le is as follows:
x1 = (0:3)’;x2 = [5,7,8,11]’;
y = [7.1,19.2,31,45]’;
X = [ones(size(x1)), x1, x2];
a = X\y
yp = X*a;
Max_Percent_Error = 100*max(abs((yp-y)./y))
The vector yp is the vector of breaking strength values predicted by the model. The scalar
Max_Percent_Error is the maximum percent error in the four predictions. The
results are a = [0.8000, 10.2429, 1.2143]’ and Max_Percent_Error =
3.2193. Thus the model is y 0.8 10.2429x1 1.2143x2. The maximum percent
error of the model’s predictions, as compared to the given data, is 3.2193 percent.
Linear-in-Parameters Regression
Sometimes we want to t an expression that is neither a polynomial nor a function that can be converted to linear form by a logarithmic or other transformation. In some cases we can still do a least-squares t if the function is a linear
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281
expression in terms of its parameters. The following example illustrates the
method.
Response of a Biomedical Instrument
EXAMPLE 6.2–4
Engineers developing instrumentation often need to obtain a response curve that describes
how fast the instrument can make measurements. The theory of instrumentation shows that
often the response can be described by one of the following equations, where ␷ is the voltage
output and t is time. In both models, the voltage reaches a steady-state constant value as
t : q , and T is the time required for the voltage to equal 95 percent of the steady-state
value.
␷(t) = a1 + a2e-3t>T
␷(t) = a1 + a2e
(first- order model)
+ a3te
-3t>T
-3t>T
(second- order model)
The following data give the output voltage of a certain device as a function of time. Obtain a function that describes these data.
t (s)
0
0.3
0.8
1.1
1.6
2.3
3
␷ (V)
0
0.6
1.28
1.5
1.7
1.75
1.8
■ Solution
Plotting the data, we estimate that it takes approximately 3 s for the voltage to become
constant. Thus we estimate that T 3. The rst-order model written for each of the n data
points results in n equations, which can be expressed as follows:
y1
1 e-t1
-t2
y2
a
1 e
≥Á Á ¥ c 1 d = ≥Á¥
a2
yn
1 e-tn
or, in matrix form,
Xa = y¿
which can be solved for the coef cient vector a using left division. The following
MATLAB script solves the problem.
t = [0,0.3,0.8,1.1,1.6,2.3,3];
y = [0,0.6,1.28,1.5,1.7,1.75,1.8];
X = [ones(size(t));exp(-t)]’;
a = X\y’
The answer is a1 2.0258 and a2 1.9307.
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2
1.8
1.6
1.4
1.2
v (V)
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Data
First Order
Second Order
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
t (s)
2
2.5
3
Figure 6.2–4 Comparison of rst- and second-order model ts.
A similar procedure can be followed for the second-order model.
1 e-t1
1 e-t2
≥Á Á
t1e-t1
t2e-t2
Á¥
1
tne-tn
e-tn
Continue the previous script as follows.
y1
a1
y2
a2 = ≥Á ¥
Ja K
3
yn
X = [ones(size(t));exp(-t);t.*exp(-t)]’;
a = X\y’
The answer is a1 1.7496, a2 1.7682, and a3 0.8885. The two models are plotted
with the data in Figure 6.2–4. Clearly the second-order model gives the better t.
6.3 The Basic Fitting Interface
MATLAB supports curve tting through the Basic Fitting interface. Using this
interface, you can quickly perform basic curve- tting tasks within the same easyto-use environment. The interface is designed so that you can do the following:
■
■
■
Fit data using a cubic spline or a polynomial up to degree 10.
Plot multiple ts simultaneously for a given data set.
Plot the residuals.
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6.3
■
■
■
■
The Basic Fitting Interface
Examine the numerical results of a t.
Interpolate or extrapolate a t.
Annotate the plot with the numerical t results and the norm of residuals.
Save the t and evaluated results to the MA TLAB workspace.
Depending on your specific curve-fitting application, you can use the Basic
Fitting interface, the command line functions, or both. Note: You can use the
Basic Fitting interface only with two-dimensional data. However, if you plot
multiple data sets as a subplot, and at least one data set is two-dimensional,
then the interface is enabled.
Two panes of the Basic Fitting interface are shown in Figure 6.3–1. To reproduce this state:
1. Plot some data.
2. Select Basic Fitting from the Tools menu of the Figure window.
3. When the rst pane of the Basic Fitting interface appears, click the right
arrow button once.
Figure 6.3–1 The Basic Fitting interface.
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The third pane is used for interpolating or extrapolating a t. It appears when you
click the right arrow button a second time.
At the top of the rst pane is the Select data window which contains the
names of all the data sets you display in the Figure window associated with the
Basic Fitting interface. Use this menu to select the data set to be t. You can
perform multiple ts for the current data set. Use the Plot Editor to change the
name of a data set. The remaining items on the rst pane are used as follows.
■
■
■
■
■
■
■
Center and scale X data. If checked, the data are centered at zero mean
and scaled to unit standard deviation. You may need to center and scale
your data to improve the accuracy of the subsequent numerical computations. As described in the previous section, a warning is returned to the
Command window if a t produces results that might be inaccurate.
Plot ts. This panel allows you to visually explore one or more ts to the
current data set.
Check to display ts on gur e. Select the ts you want to display for the
current data set. You can choose as many ts for a given data set as you want.
However, if your data set has n points, then you should use polynomials with
at most n coef cients. If you t using polynomials with more than n
coef cients, the interface will automatically set a suf cient number of
coef cients to zero during the calculation so that a solution can be obtained.
Show equations. If checked, the t equation is displayed on the plot.
Signi cant digits. Select the signi cant digits associated with the t
coef cient display .
Plot residuals. If checked, the residuals are displayed. You can display the
residuals as a bar plot, a scatter plot, a line plot using either the same gure
window as the data or using a separate gure window . If you plot multiple
data sets as a subplot, then residuals can be plotted only in a separate gure
window. See Figure 6.3–2.
Show norm of residuals. If checked, the norm of residuals is displayed. The
norm of residuals is a measure of the goodness of t, where a smaller value
indicates a better t. The norm is the square root of the sum of the squares of
the residuals.
The second pane of the Basic Fitting interface is labeled Numerical Results. This
pane enables you to explore the numerical results of a single t to the current data
set without plotting the t. It contains three items.
■
■
Fit. Use this menu to select an equation to t to the current data set. The t
results are displayed in the box below the menu. Note that selecting an
equation in this menu does not affect the state of the Plot ts selection.
Therefore, if you want to display the t in the data plot, you might need to
check the relevant check box in Plot ts.
Coef cients and norm of r esiduals. Displays the numerical results for the
equation selected in Fit. Note that when you rst open the Numerical
Results panel, the results of the last t you selected in Plot ts are displayed.
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6.4
Summary
Figure 6.3–2 A gure produced by the Basic Fitting interface.
■
Save to workspace. Launches a dialog box that allows you to save the t
results to workspace variables.
The third pane of the Basic Fitting interface contains three items.
■
■
■
Find Y ⴝ f(X). Use this to interpolate or extrapolate the current fit.
Enter a scalar or a vector of values corresponding to the independent
variable (X). The current fit is evaluated after you click on the Evaluate
button, and the results are displayed in the associated window. The
current fit is displayed in the Fit window.
Save to workspace. Launches a dialog box that allows you to save the
evaluated results to workspace variables.
Plot evaluated results. If checked, the evaluated results are displayed on
the data plot.
6.4 Summary
In this chapter you learned an important application of plotting—function
discovery—which is the technique for using data plots to obtain a mathematical
function that describes the data. Regression can be used to develop a model for
cases where there is considerable scatter in the data.
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Many physical processes can be modeled with functions that produce a
straight line when plotted using a suitable set of axes. In some cases, we can nd
a transformation that produces a straight line in the transformed variable.
When such a function or transformation cannot be found, we resort to polynomial regression, multiple linear regression, or linear-in-parameters regression
to obtain an approximate functional description of the data. The MATLAB Basic
Fitting interface is a powerful aid in obtaining regression models.
Key Terms with Page References
Coef cient of determination, 275
Linear-in-parameters, 280
Multiple linear regression, 279
Regression, 271
Residuals, 272
Problems
You can nd the answers to problems marked with an asterisk at the end of the text.
Section 6.1
1.
The distance a spring stretches from its “free length” is a function of how
much tension force is applied to it. The following table gives the spring
length y that the given applied force f produced in a particular spring. The
spring’s free length is 4.7 in. Find a functional relation between f and x,
the extension from the free length (x y 4.7).
Force f (lb)
Spring length y (in.)
0
0.94
2.30
3.28
4.7
7.2
10.6
12.9
2.* In each of the following problems, determine the best function y(x)
(linear, exponential, or power function) to describe the data. Plot the
function on the same plot with the data. Label and format the plots
appropriately.
a.
x
25
30
35
40
45
5
y
260
480
745
1100
b.
x
2.5
3.5
4
4.5
5
5.5
6
7
8
9
10
y
1500 1220 1050
3
915
810
745
690
620
520
480
410
390
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Problems
c.
3.
x
550
600
650
700
750
y
41.2 18.62
8.62
3.92
1.86
The population data for a certain country are as follows:
Year
2004
2005
2006
2007
2008
2009
Population (millions)
10
10.9
11.7
12.6
13.8
14.9
Obtain a function that describes these data. Plot the function and the
data on the same plot. Estimate when the population will be double its
2004 size.
4.* The half-life of a radioactive substance is the time it takes to decay by
one-half. The half-life of carbon 14, which is used for dating previously
living things, is 5500 years. When an organism dies, it stops accumulating
carbon 14. The carbon 14 present at the time of death decays with time.
Let C(t)/C(0) be the fraction of carbon 14 remaining at time t. In
radioactive carbon dating, scientists usually assume that the remaining
fraction decays exponentially according to the following formula:
C(t)
= e-bt
C(0)
a. Use the half-life of carbon 14 to nd the value of the parameter b, and
plot the function.
b. If 90 percent of the original carbon 14 remains, estimate how long ago
the organism died.
c. Suppose our estimate of b is off by 1 percent. How does this error
affect the age estimate?
5.
6.
Quenching is the process of immersing a hot metal object in a bath for a
speci ed time to obtain certain properties such as hardness. A copper sphere
25 mm in diameter, initially at 300C, is immersed in a bath at 0C. The following table gives measurements of the sphere’s temperature versus time.
Find a functional description of these data. Plot the function and the data on
the same plot.
Time (s)
0
1
2
3
4
5
6
Temperature (C)
300
150
75
35
12
5
2
The useful life of a machine bearing depends on its operating temperature, as
the following data show. Obtain a functional description of these data. Plot
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the function and the data on the same plot. Estimate a bearing’s life if it
operates at 150F.
Temperature (F)
Bearing life (hours
7.
103)
100
120
140
160
180
200
220
28
21
15
11
8
6
4
A certain electric circuit has a resistor and a capacitor. The capacitor is
initially charged to 100 V. When the power supply is detached, the
capacitor voltage decays with time, as the following data table shows.
Find a functional description of the capacitor voltage ␷ as a function of
time t. Plot the function and the data on the same plot.
Time (s)
0
0.5
1
1.5
2
2.5
3
3.5
4
Voltage (V)
100
62
38
21
13
7
4
2
3
Sections 6.2 and 6.3
8.* The distance a spring stretches from its free length is a function of how
much tension force is applied to it. The following table gives the spring
length y that was produced in a particular spring by the given applied
force f. The spring’s free length is 4.7 in. Find a functional relation
between f and x, the extension from the free length (x y 4.7).
9.
Force f (lb)
Spring length y (in.)
0
0.94
2.30
3.28
4.7
7.2
10.6
12.9
The following data give the drying time T of a certain paint as a function
of the amount of a certain additive A.
a. Find the rst-, second-, third-, and fourth-degree polynomials that t
the data, and plot each polynomial with the data. Determine the quality
of the curve t for each by computing J, S, and r 2.
b. Use the polynomial giving the best t to estimate the amount of
additive that minimizes the drying time.
A (oz)
0
1
2
3
4
5
6
7
8
9
T (min)
130
115
110
90
89
89
95
100
110
125
10.* The following data give the stopping distance d as a function of initial
speed ␷, for a certain car model. Find a quadratic polynomial that ts the
data. Determine the quality of the curve t by computing J, S, and r2.
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Problems
␷ (mi/hr)
20
30
40
50
60
70
d (ft)
45
80
130
185
250
330
11.* The number of twists y required to break a certain rod is a function of the
percentage x1 and x2 of each of two alloying elements present in the rod. The
following table gives some pertinent data. Use linear multiple regression to
obtain a model y a0 a1x1 a2x2 of the relationship between the number
of twists and the alloy percentages. In addition, nd the maximum percent
error in the predictions.
Number of twists
y
Percentage of element 1
x1
Percentage of element 2
x2
40
51
65
72
38
46
53
67
31
39
48
56
1
2
3
4
1
2
3
4
1
2
3
4
1
1
1
1
2
2
2
2
3
3
3
3
12.
The following represents pressure samples, in pounds per square inch
(psi), taken in a fuel line once every second for 10 sec.
Time (sec)
Pressure (psi)
Time (sec)
Pressure (psi)
1
2
3
4
5
26.1
27.0
28.2
29.0
29.8
6
7
8
9
10
30.6
31.1
31.3
31.0
30.5
a. Fit a rst-degree polynomial, a second-degree polynomial, and a
third-degree polynomial to these data. Plot the curve ts along with the
data points.
b. Use the results from part a to predict the pressure at t 11 sec.
Explain which curve t gives the most reliable prediction. Consider
the coef cients of determination and the residuals for each t in making your decision.
13.
A liquid boils when its vapor pressure equals the external pressure acting
on the surface of the liquid. This is why water boils at a lower temperature at higher altitudes. This information is important for people who
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must design processes utilizing boiling liquids. Data on the vapor pressure P of water as a function of temperature T are given in the following
table. From theory we know that ln P is proportional to 1/T . Obtain a
curve t for P(T) from these data. Use the t to estimate the vapor
pressure at 285 and 300 K.
14.
T (K)
P (torr)
273
278
283
288
293
298
4.579
6.543
9.209
12.788
17.535
23.756
The solubility of salt in water is a function of the water temperature. Let S
represent the solubility of NaCl (sodium chloride) as grams of salt in 100 g
of water. Let T be temperature in C. Use the following data to obtain a
curve t for S as a function of T. Use the t to estimate S when T 25C.
T (C) S (g NaCl/100 g H2O)
10
20
30
40
50
60
70
80
90
15.
35
35.6
36.25
36.9
37.5
38.1
38.8
39.4
40
The solubility of oxygen in water is a function of the water temperature.
Let S represent the solubility of O2 as millimoles of O2 per liter of
water. Let T be temperature in C. Use the following data to obtain a
curve t for S as a function of T. Use the t to estimate S when
T 8C and T 50C.
T (C)
S (mmol O2 /L H2O)
5
10
15
20
25
30
35
40
45
1.95
1.7
1.55
1.40
1.30
1.15
1.05
1.00
0.95
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Problems
16.
The following function is linear in the parameters a1 and a2.
y(x) = a1 + a2 ln x
Use least-squares regression with the following data to estimate the
values of a1 and a2. Use the curve t to estimate the values of y at x 2.5
and at x 11.
17.
x
1
2
3
4
5
6
7
8
9
10
y
10
14
16
18
19
20
21
22
23
23
Chemists and engineers must be able to predict the changes in chemical
concentration in a reaction. A model used for many single-reactant
processes is
Rate of change of concentration = - kCn
where C is the chemical concentration and k is the rate constant. The
order of the reaction is the value of the exponent n. Solution methods for
differential equations (which are discussed in Chapter 9) can show that
the solution for a rst-order reaction ( n 1) is
C(t) = C(0)e-kt
The following data describe the reaction
(CH3)3CBr + H2O : (CH3)3COH + HBr
Use these data to obtain a least-squares t to estimate the value of k.
18.
Time t (h)
C(mol of (CH3)3 CBr/L)
0
3.15
6.20
10.0
18.3
30.8
43.8
0.1039
0.0896
0.0776
0.0639
0.0353
0.0207
0.0101
Chemists and engineers must be able to predict the changes in chemical
concentration in a reaction. A model used for many single-reactant
processes is
Rate of change of concentration = - kCn
where C is the chemical concentration and k is the rate constant. The
order of the reaction is the value of the exponent n. Solution methods for
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CHAPTER 6 Model Building and Regression
differential equations (which are discussed in Chapter 9) can show that
the solution for a rst-order reaction ( n 1) is
C(t) = C(0)e-kt
and the solution for a second-order reaction (n 2) is
1
1
=
+ kt
C(t)
C(0)
The following data (from Brown, 1994) describes the gas-phase
decomposition of nitrogen dioxide at 300C.
2NO2 : 2NO + O2
Time t (s)
C (mol NO2 /L)
0
50
100
200
300
0.0100
0.0079
0.0065
0.0048
0.0038
Determine whether this is a rst-order or second-order reaction, and
estimate the value of the rate constant k.
19.
Chemists and engineers must be able to predict the changes in
chemical concentration in a reaction. A model used for many singlereactant processes is
Rate of change of concentration = - kCn
where C is the chemical concentration and k is the rate constant. The
order of the reaction is the value of the exponent n. Solution methods for
differential equations (which are discussed in Chapter 9) can show that
the solution for a rst-order reaction ( n 1) is
C(t) = C(0)e-kt
The solution for a second-order reaction (n 2) is
1
1
=
+ kt
C(t)
C(0)
and the solution for a third-order reaction (n 3) is
1
1
2
2C (t)
=
2C2(0)
+ kt
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Problems
Time t (min)
C (mol of reactant/L)
5
10
15
20
25
30
35
40
45
0.3575
0.3010
0.2505
0.2095
0.1800
0.1500
0.1245
0.1070
0.0865
The preceding data describe a certain reaction. By examining the
residuals, determine whether this is a rst-order , second-order, or
third-order reaction, and estimate the value of the rate constant k.
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© The McGraw-Hill Companies, Inc./Mark Dierker, Photographer.
Engineering in the
21st Century
Energy-Efficient Transportation
M
odern societies have become very dependent on transportation powered by gasoline and diesel fuel. There is some disagreement about
how long it will take to exhaust these fuel resources, but it will certainly happen. Novel engineering developments in both personal and mass transportation will be needed to reduce our dependence on such fuels. These
developments will be required in a number of areas such as engine design, electric motor and battery technology, lightweight materials, and aerodynamics.
A number of such initiatives are underway. Several projects have the goal of
designing a six-passenger car that is one-third lighter and 40 percent more aerodynamic than today’s sleekest cars. A hybrid gas-electric vehicle is the most promising at present. An internal combustion engine and an electric motor drive the
wheels. A fuel cell or a battery is charged either by a generator driven by the engine
or by energy recovered during braking. This is called regenerative braking.
The weight reduction can be achieved with all-aluminum unibody construction and by improved design of the engine, radiator, and brakes to make use of
advanced materials such as composites and magnesium. Other manufacturers are
investigating plastic bodies made from recycled materials.
There is still much room for improved ef ciency , and research and development engineers in this area will remain busy for some time. MATLAB is widely
used to assist these efforts with modeling and analysis tools for hybrid vehicle
designs. ■
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C H A P T E R
7
Statistics, Probability,
and Interpolation
OUTLINE
7.1 Statistics and Histograms
7.2 The Normal Distribution
7.3 Random Number Generation
7.4 Interpolation
7.5 Summary
Problems
This chapter begins with an introduction to basic statistics in Section 7.1. You
will see how to obtain and interpret histograms, which are specialized plots
for displaying statistical results. The normal distribution, commonly called
the bell-shaped curve, forms the basis of much of probability theory and many
statistical methods. It is covered in Section 7.2. In Section 7.3 you will see
how to include random processes in your simulation programs. In Section 7.4
you will see how to use interpolation with data tables to estimate values that
are not in the table.
When you have nished this chapter , you should be able to use MATLAB to
do the following:
■
■
■
Solve basic problems in statistics and probability.
Create simulations incorporating random processes.
Apply interpolation techniques.
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7.1 Statistics and Histograms
MEAN
MODE
MEDIAN
BINS
With MATLAB you can compute the mean (the average), the mode (the most
frequently occurring value), and the median (the middle value) of a set of data.
MATLAB provides the mean(x), mode(x), and median(x) functions to
compute the mean, mode, and median of the data values stored in x, if x is a vector. However, if x is a matrix, a row vector is returned containing the mean (or
mode or median) value of each column of x. These functions do not require the
elements in x to be sorted in ascending or descending order.
The way the data are spread around the mean can be described by a histogram plot. A histogram is a plot of the frequency of occurrence of data values
versus the values themselves. It is a bar plot of the number of data values that
occur within each range, with the bar centered in the middle of the range.
To plot a histogram, you must group the data into subranges, called bins. The
choice of the bin width and bin center can drastically change the shape of the
histogram. If the number of data values is relatively small, the bin width cannot
be small because some of the bins will contain no data and the resulting histogram might not usefully illustrate the distribution of the data.
To obtain a histogram, rst sort the data values if they have has not yet been
sorted (you can use the sort function here). Then choose the bin ranges and bin
centers and count the number of values in each bin. Use the bar function to plot
the number of values in each bin versus the bin centers as a bar chart. The function bar(x,y) creates a bar chart of y versus x.
MATLAB also provides the hist command to generate a histogram.
This command has several forms. Its basic form is hist(y), where y is a
vector containing the data. This form aggregates the data into 10 bins evenly
spaced between the minimum and maximum values in y. The second form is
hist(y,n), where n is a user-speci ed scalar indicating the number of bins.
The third form is hist(y,x), where x is a user-speci ed vector that determines the location of the bin centers; the bin widths are the distances between
the centers.
Breaking Strength of Thread
EXAMPLE 7.1–1
To ensure proper quality control, a thread manufacturer selects samples and tests them for
breaking strength. Suppose that 20 thread samples are pulled until they break, and the
breaking force is measured in newtons rounded off to integer values. The breaking force
values recorded were 92, 94, 93, 96, 93, 94, 95, 96, 91, 93, 95, 95, 95, 92, 93, 94, 91, 94,
92, and 93. Plot the histogram of the data.
■ Solution
Store the data in the vector y, which is shown in the following script le. Because there
are six outcomes (91, 92, 93, 94, 95, 96 N), we choose six bins. However, if you use
hist(y,6), the bins will not be centered at 91, 92, 93, 94, 95, and 96. So use the form
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297
Absolute Frequency Histogram for 20 Tests
6
Absolute Frequency
5
4
3
2
1
0
90
91
92
93
94
Thread Strength (N)
95
96
97
Figure 7.1–1 Histograms for 20 tests of thread strength.
hist(y,x), where x = 91:96. The following script le generates the histogram
shown in Figure 7.1–1.
% Thread breaking strength data for 20 tests.
y = [92,94,93,96,93,94,95,96,91,93,...
95,95,95,92,93,94,91,94,92,93];
% The six possible outcomes are 91,92,93,94,95,96.
x = 91:96;
hist(y,x),axis([90 97 0 6]),ylabel(‘Absolute Frequency’),...
xlabel(‘Thread Strength (N)’),...
title(‘Absolute Frequency Histogram for 20 Tests’)
The absolute frequency is the number of times a particular outcome occurs. For example, in 20 tests these data show that a 95 occurred 4 times. The
absolute frequency is 4, and its relative frequency is 4/20, or 20 percent of the
time.
When there is a large amount of data, you can avoid typing in every data
value by rst aggregating the data. The following example shows how this is done
using the ones function. The following data were generated by testing 100 thread
ABSOLUTE
FREQUENCY
RELATIVE
FREQUENCY
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Absolute Frequency Histogram for 100 Tests
25
20
Absolute Frequency
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10
5
0
90
91
92
93
94
Thread Strength (N)
95
96
97
Figure 7.1–2 Absolute frequency histogram for 100 thread tests.
samples. The number of times 91, 92, 93, 94, 95, or 96 N was measured is 13, 15,
22, 19, 17, and 14, respectively.
% Thread strength data for 100 tests.
y = [91*ones(1,13),92*ones(1,15),93*ones(1,22),...
94*ones(1,19),95*ones(1,17),96*ones(1,14)];
x = 91:96;
hist(y,x),ylabel(‘Absolute Frequency’),...
xlabel(‘Thread Strength (N)’),...
title(‘Absolute Frequency Histogram for 100 Tests’)
The result appears in Figure 7.1–2.
The hist function is somewhat limited in its ability to produce useful histograms. Unless all the outcome values are the same as the bin centers (as is the
case with the thread examples), the graph produced by the hist function will
not be satisfactory. This case occurs when you want to obtain a relative
frequency histogram. In such cases you can use the bar function to generate the
histogram. The following script le generates the relative frequency histogram
for the 100 thread tests. Note that if you use the bar function, you must aggregate the data rst.
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% Relative frequency histogram using the bar function.
tests = 100;
y = [13,15,22,19,17,14]/tests;
x = 91:96;
bar(x,y),ylabel(‘Relative Frequency’),...
xlabel(‘Thread Strength (N)’),...
title(‘Relative Frequency Histogram for 100 Tests’)
The result appears in Figure 7.1–3.
The fourth, fth, and sixth forms of the hist function do not generate a
plot, but are used to compute the frequency counts and bin locations. The bar
function can then be used to plot the histogram. The syntax of the fourth form
is [z,x] = hist(y), where z is the returned vector containing the frequency
count and x is the returned vector containing the bin locations. The fth and sixth
forms are [z,x] = hist(y,n) and [z,x] = hist(y,x). In the latter
case the returned vector x is the same as the user-supplied vector. The following
script le shows how the sixth form can be used to generate a relative frequency
histogram for the thread example with 100 tests.
tests = 100;
y = [91*ones(1,13),92*ones(1,15),93*ones(1,22),...
94*ones(1,19),95*ones(1,17),96*ones(1,14);
Relative Frequency Histogram for 100 Tests
0.25
Relative Frequency
0.2
0.15
0.1
0.05
0
90
91
92
93
94
Thread Strength (N)
95
Figure 7.1–3 Relative frequency histogram for 100 thread tests.
96
97
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Table 7.1–1 Histogram functions
Command
Description
bar(x,y)
hist(y)
Creates a bar chart of y versus x.
Aggregates the data in the vector y into 10 bins evenly spaced
between the minimum and maximum values in y.
Aggregates the data in the vector y into n bins evenly spaced
between the minimum and maximum values in y.
Aggregates the data in the vector y into bins whose center
locations are speci ed by the vector x. The bin widths are the
distances between the centers.
Same as hist(y) but returns two vectors z and x that
contain the frequency count and the bin locations.
Same as hist(y,n) but returns two vectors z and x that
contain the frequency count and the bin locations.
Same as hist(y,x) but returns two vectors z and x that
contain the frequency count and the bin locations. The
returned vector x is the same as the user-supplied vector x.
hist(y,n)
hist(y,x)
[z,x] = hist(y)
[z,x] = hist(y,n)
[z,x] = hist(y,x)
x = 91:96;
[z,x] = hist(y,x);bar(x,z/tests),...
ylabel(‘Relative Frequency’),xlabel(‘Thread Strength(N)’),...
title(‘Relative Frequency Histogram for 100 Tests’)
The plot generated by this M- le will be identical to that shown in Figure 7.1–3.
These commands are summarized in Table 7.1–1.
Test Your Understanding
T7.1–1 In 50 tests of thread, the number of times 91, 92, 93, 94, 95, or 96 N was
measured was 7, 8, 10, 6, 12, and 7, respectively. Obtain the absolute
and relative frequency histograms.
The Data Statistics Tool
With the Data Statistics tool you can calculate statistics for data and add plots
of the statistics to a graph of the data. The tool is accessed from the Figure
window after you plot the data. Click on the Tools menu, then select Data
Statistics. The menu appears as shown in Figure 7.1–4. To show the mean of
the dependent variable (y) on the plot, click the box in the row labeled mean
under the column labeled Y, as shown in the gure. A horizontal line is then
placed on the plot at the mean. You can plot other statistics as well; these are
shown in the gure. You can save the statistics to the workspace as a structure
by clicking on the Save to Workspace button. This opens a dialog box that
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Figure 7.1–4 The Data Statistics tool.
prompts you for a name for the structure containing the x data, and a name for
the y data structure.
7.2 The Normal Distribution
Rolling a die is an example of a process whose possible outcomes are a limited
set of numbers, namely, the integers from 1 to 6. For such processes the probability is a function of a discrete-valued variable, that is, a variable having a limited number of values. For example, Table 7.2–1 gives the measured heights of
100 men 20 years of age. The heights were recorded to the nearest 1/2 in., so the
height variable is discrete-valued.
Scaled Frequency Histogram
You can plot the data as a histogram using either the absolute or relative
frequencies. However, another useful histogram uses data scaled so that the
total area under the histogram’s rectangles is 1. This scaled frequency histogram is the absolute frequency histogram divided by the total area of that
histogram. The area of each rectangle on the absolute frequency histogram
equals the bin width times the absolute frequency for that bin. Because all the
rectangles have the same width, the total area is the bin width times the sum
of the absolute frequencies. The following M-file produces the scaled histogram shown in Figure 7.2–1.
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Table 7.2–1 Height data for men 20 years of age
Height (in.)
Frequency
Height (in.)
Frequency
64
64.5
65
65.5
66
66.5
67
67.5
68
68.5
69
69.5
1
0
0
0
2
4
5
4
8
11
12
10
70
70.5
71
71.5
72
72.5
73
73.5
74
74.5
75
9
8
7
5
4
4
3
1
1
0
1
% Absolute frequency data.
y_abs=[1,0,0,0,2,4,5,4,8,11,12,10,9,8,7,5,4,4,3,1,1,0,1];
binwidth = 0.5;
% Compute scaled frequency data.
area = binwidth*sum(y_abs);
y_scaled = y_abs/area;
0.25
0.2
Scaled Frequency
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0.15
0.1
0.05
0
62
64
66
68
70
Height (in.)
Figure 7.2–1 Scaled histogram of height data.
72
74
76
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% De ne the bins.
bins = 64:binwidth:75;
% Plot the scaled histogram.
bar(bins,y_scaled),...
ylabel(‘Scaled Frequency’),xlabel(‘Height (in.)’)
Because the total area under the scaled histogram is 1, the fractional area
corresponding to a range of heights gives the probability that a randomly selected
20-year-old man will have a height in that range. For example, the heights of the
scaled histogram rectangles corresponding to heights of 67 through 69 in. are
0.1, 0.08, 0.16, 0.22, and 0.24. Because the bin width is 0.5, the total area corresponding to these rectangles is (0.1⫹ 0.08 ⫹ 0.16 ⫹ 0.22 ⫹ 0.24)(0.5) ⫽ 0.4.
Thus 40 percent of the heights lie between 67 and 69 in.
You can use the cumsum function to calculate areas under the scaled
frequency histogram and therefore to calculate probabilities. If x is a vector,
cumsum(x) returns a vector the same length as x, whose elements are the sum
of the previous elements. For example, if x =[2, 5, 3, 8], cumsum(x)=
[2, 7, 10, 18]. If A is a matrix, cumsum(A) computes the cumulative sum
of each row. The result is a matrix the same size as A.
After running the previous script, the last element of cumsum(y_scaled)*
binwidth is 1, which is the area under the scaled frequency histogram. To
compute the probability of a height lying between 67 and 69 in. (that is, above the
6th value up to the 11th value), type
>>prob = cumsum(y_scaled)*binwidth;
>>prob67_69 = prob(11)-prob(6)
The result is prob67_69 = 0.4000, which agrees with our previous
calculation of 40 percent.
Continuous Approximation to the Scaled Histogram
For processes having an in nite number of possible outcomes, the probability
is a function of a continuous variable and is plotted as a curve rather than as
rectangles. It is based on the same concept as the scaled histogram; that is, the
total area under the curve is 1, and the fractional area gives the probability of
occurrence of a speci c range of outcomes. A probability function that
describes many processes is the normal or Gaussian function, which is shown
in Figure 7.2–2.
This function is also known as the bell-shaped curve. Outcomes that can be
described by this function are said to be normally distributed. The normal
probability function is a two-parameter function; one parameter, ␮, is the mean
of the outcomes, and the other parameter, ␴, is the standard deviation. The mean
␮ locates the peak of the curve and is the most likely value to occur. The width,
or spread, of the curve is described by the parameter ␴. Sometimes the term
variance is used to describe the spread of the curve. The variance is the square of
the standard deviation ␴.
NORMAL OR
GAUSSIAN
FUNCTION
NORMALLY
DISTRIBUTED
STANDARD
DEVIATION
VARIANCE
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p
1/σ√2
σ π
2σ
0.6065/σ
σ √2π
x
μ +– σ
μ
μ+σ
Figure 7.2–2 The basic shape of the normal distribution curve.
The normal probability function is described by the following equation:
p(x) =
1
2
2
e-(x-␮) >2␴
␴ 12␲
(7.2–1)
It can be shown that approximately 68 percent of the area lies between the
limits of ␮ ⫺ ␴ ⱕ x ⱕ ␮ ⫹ ␴. Consequently, if a variable is normally distributed,
there is a 68 percent chance that a randomly selected sample will lie within one
standard deviation of the mean. In addition, approximately 96 percent of the area
lies between the limits of ␮ ⫺ 2␴ ⱕ x ⱕ ␮ ⫹ 2␴, and 99.7 percent, or practically
100 percent, of the area lies between the limits of ␮ ⫺ 3␴ ⱕ x ⱕ ␮ ⫹ 3␴.
The functions mean(x), var(x), and std(x) compute the mean, variance, and standard deviation of the elements in the vector x.
Mean and Standard Deviation of Heights
EXAMPLE 7.2–1
Statistical analysis of data on human proportions is required in many engineering
applications. For example, designers of submarine crew quarters need to know how small
they can make bunk lengths without eliminating a large percentage of prospective crew
members. Use MATLAB to estimate the mean and standard deviation for the height data
given in Table 7.2–1.
■ Solution
The script le follows. The data given in Table 7.2–1 are the absolute frequency data and
are stored in the vector y_abs. A bin width of 1/2 in. is used because the heights were
measured to the nearest 1/2 in. The vector bins contains the heights in 1/2 in. increments.
To compute the mean and standard deviation, reconstruct the original (raw) height
data from the absolute frequency data. Note that these data have some zero entries. For
example, none of the 100 men had a height of 65 in. Thus to reconstruct the raw data, start
with an empty vector y_raw and ll it with the height data obtained from the absolute
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frequencies. The for loop checks to see whether the absolute frequency for a particular
bin is nonzero. If it is nonzero, append the appropriate number of data values to the
vector y_raw. If the particular bin frequency is 0, y_raw is left unchanged.
% Absolute frequency data.
y_abs = [1,0,0,0,2,4,5,4,8,11,12,10,9,8,7,5,4,4,3,1,1,0,1];
binwidth = 0.5;
% De ne the bins.
bins = [64:binwidth:75];
% Fill the vector y_raw with the raw data.
% Start with an empty vector.
y_raw = [];
for i = 1:length(y_abs)
if y_abs(i)>0
new = bins(i)*ones(1,y_abs(i));
else
new = [];
end
y_raw = [y_raw,new];
end
% Compute the mean and standard deviation.
mu = mean(y_raw),sigma = std(y_raw)
When you run this program, you will nd that the mean is ␮ ⫽ 69.6 in. and the standard deviation is ␴ ⫽ 1.96 in.
If you need to compute probabilities based on the normal distribution, you
can use the erf function. Typing erf(x) returns the area to the left of the value
2
t ⫽ x under the curve of the function 2e-t > 1␲. This area, which is a function of x,
is known as the error function and is written as erf(x). The probability that the
random variable x is less than or equal to b is written as P(x ⱕ b) if the outcomes
are normally distributed. This probability can be computed from the error function
as follows:
P(x … b) =
b - ␮
1
c 1 + erf a
bd
2
␴ 12
ERROR FUNCTION
(7.2–2)
The probability that the random variable x is no less than a and no greater than b
is written as P(a ⱕ x ⱕ b). It can be computed as follows:
P(a … x … b) =
a - ␮
b - ␮
1
c erf a
b - erf a
bd
2
␴ 12
␴ 12
(7.2–3)
Estimation of Height Distribution
Use the results of Example 7.2–1 to estimate how many 20-year-old men are no taller
than 68 in. How many are within 3 in. of the mean?
EXAMPLE 7.2–2
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■ Solution
In Example 7.2–1 the mean and standard deviation were found to be ␮ ⫽ 69.3 in. and
␴ ⫽ 1.96 in. In Table 7.2–1, note that few data points are available for heights less than
68 in. However, if you assume that the heights are normally distributed, you can use
Equation (7.2–2) to estimate how many men are shorter than 68 in. Use (7.2–2) with
b ⫽ 68, that is,
P(x … 68) =
68 - 69.3
1
bd
c 1 + erf a
2
1.96 12
To determine how many men are within 3 in. of the mean, use Equation (7.2–3) with a ⫽
␮ ⫺ 3 ⫽ 66.3 and b ⫽ ␮ ⫹ 3 ⫽ 72.3, that is,
P(66.3 … x … 72.3) =
1
-3
3
b - erf a
bd
c erf a
2
1.96 12
1.96 12
In MATLAB these expressions are computed in a script le as follows:
mu = 69.3;
s = 1.96;
% How many are no taller than 68 inches?
b1 = 68;
P1 = (1+erf((b1-mu)/(s*sqrt(2))))/2
% How many are within 3 inches of the mean?
a2 = 66.3;
b2 = 72.3;
P2 = (erf((b2-mu)/(s*sqrt(2)))-erf((a2-mu)/(s*sqrt(2))))/2
When you run this program, you obtain the results P1 = 0.2536 and P2 = 0.8741.
Thus 25 percent of 20-year-old men are estimated to be 68 in. or less in height, and
87 percent are estimated to be between 66.3 and 72.3 in. tall.
Test Your Understanding
T7.2–1 Suppose that 10 more height measurements are obtained so that the following numbers must be added to Table 7.2–1.
Height (in.)
Additional data
64.5
65
66
67.5
70
73
74
1
2
1
2
2
1
1
(a) Plot the scaled frequency histogram. (b) Find the mean and standard
deviation. (c) Use the mean and standard deviation to estimate how many
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20-year-old men are no taller than 69 in. (d) Estimate how many are
between 68 and 72 in. tall.
(Answers: (b) mean ⫽ 69.4 in., standard deviation ⫽ 2.14 in.; (c) 43 percent; (d) 63 percent.)
Sums and Differences of Random Variables
It can be proved that the mean of the sum (or difference) of two independent
normally distributed random variables equals the sum (or difference) of their
means, but the variance is always the sum of the two variances. That is, if x and
y are normally distributed with means ␮x and ␮y and variances ␴2x and ␴2y, and
if u ⫽ x ⫹ y and ␷ ⫽ x ⫺ y, then
␮u = ␮x + ␮y
(7.2–4)
␮y = ␮x - ␮y
(7.2–5)
␴2u = ␴2y = ␴2x + ␴2y
(7.2–6)
These properties are applied in some of the homework problems.
7.3 Random Number Generation
We often do not have a simple probability distribution to describe the distribution
of outcomes in many engineering applications. For example, the probability that
a circuit consisting of many components will fail is a function of the number and
the age of the components, but we often cannot obtain a function to describe the
failure probability. In such cases we often resort to simulation to make predictions. The simulation program is executed many times, using a random set of
numbers to represent the failure of one or more components, and the results are
used to estimate the desired probability.
Uniformly Distributed Numbers
In a sequence of uniformly distributed random numbers, all values within a given
interval are equally likely to occur. The MATLAB function rand generates random
numbers uniformly distributed over the interval [0,1]. Type rand to obtain a single
random number in the interval [0,1]. Typing rand again generates a different number because the MATLAB algorithm used for the rand function requires a “state”
to start. MATLAB obtains this state from the computer’s CPU clock. Thus every
time the rand function is used, a different result will be obtained. For example,
rand
ans =
0.6161
rand
ans =
0.5184
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Type rand(n) to obtain an n ⫻ n matrix of uniformly distributed random
numbers in the interval [0, 1]. Type rand(m,n) to obtain an m ⫻ n matrix of
random numbers. For example, to create a 1 ⫻ 100 vector y having 100 random
values in the interval [0, 1], type y = rand(1,100). Using the rand
function this way is equivalent to typing rand 100 times. Even though there is
a single call to the rand function, the rand function’s calculation has the
effect of using a different state to obtain each of the 100 numbers so that they
will be random.
Use Y = rand(m,n,p,...) to generate a multidimensional array Y
having random elements. Typing rand(size(A)) produces an array of
random entries that is the same size as A.
For example, the following script makes a random choice between two
equally probable alternatives.
if rand < 0.5
disp(‘heads’)
else
disp(‘tails’)
end
To compare the results of two or more simulations, sometimes you will need
to generate the same sequence of random numbers each time the simulation
runs. To generate the same sequence, you must use the same state each time.
The current state s of the uniform number generator can be obtained by typing
s = rand(‘twister’). This returns a vector containing the current state
of the uniform generator. To set the state of the generator to s, type
rand(‘twister’,s). Typing rand(‘twister’,0) resets the generator
to its initial state. Typing rand(‘twister’,j), for integer j, resets the
generator to state j. Typing rand(‘twister’,sum(100*clock)) resets
the generator to a different state each time. Table 7.3–1 summarizes these
functions.
The name ‘twister’ refers to the speci c algorithm used by MA TLAB to
generate random numbers. In MATLAB Version 4, ‘seed’ was used instead of
‘twister’. In Versions 5 through 7.3, ‘state’ was used. Use ‘twister’
in Version 7.4 and later. The following session shows how to obtain the same
sequence every time rand is called.
>>rand(‘twister’,0)
>>rand
ans =
0.5488
>>rand
ans =
0.7152
>>rand(‘twister’,0)
>>rand
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Table 7.3–1 Random number functions
Command
Description
rand
Generates a single uniformly distributed random number between
0 and 1.
Generates an n ⫻ n matrix containing uniformly distributed random
numbers between 0 and 1.
Generates an m ⫻ n matrix containing uniformly distributed random
numbers between 0 and 1.
Returns a vector s containing the current state of the uniformly
distributed generator.
Sets the state of the uniformly distributed generator to s.
Resets the uniformly distributed generator to its initial state.
Resets the uniformly distributed generator to state j, for integer j.
Resets the uniformly distributed generator to a different state each
time it is executed.
Generates a single normally distributed random number having a
mean of 0 and a standard deviation of 1.
Generates an n ⫻ n matrix containing normally distributed random
numbers having a mean of 0 and a standard deviation of 1.
Generates an m ⫻ n matrix containing normally distributed random
numbers having a mean of 0 and a standard deviation of 1.
Like rand(‘state’) but for the normally distributed generator.
Like rand(‘state’,s) but for the normally distributed generator.
Like rand(‘state’,0) but for the normally distributed generator.
Like rand(‘state’,j) but for the normally distributed generator.
Like rand(‘state’,sum(100*clock)) but for the normally
distributed generator.
Generates a random permutation of the integers from 1 to n.
rand(n)
rand(m,n)
s = rand(‘state’)
rand(‘twister’,s)
rand(‘twister’,0)
rand(‘twister’,j)
rand(‘twister’,sum(100*clock))
randn
randn(n)
randn(m,n)
s = randn(‘state’)
randn(‘state’,s)
randn(‘state’,0)
randn(‘state’,j)
randn(‘state’,sum(100*clock))
randperm(n)
ans =
0.5488
>>rand
ans =
0.7152
You need not start with the initial state to generate the same sequence. To show
this, continue the above session as follows.
>>s = rand(‘twister’);
>>rand(‘twister’,s)
>>rand
ans =
0.6028
>>rand(‘twister’,s)
>>rand
ans =
0.6028
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You can use the rand function to generate random numbers in an interval
other than [0, 1]. For example, to generate values in the interval [2,10], generate
a random number between 0 and 1, multiply it by 8 (the difference between
the upper and lower bounds), and add the lower bound (2). The result is a value
that is uniformly distributed in the interval [2,10]. The general formula for
generating a uniformly distributed random number y in the interval [a,b] is
y = (b - a)x + a
(7.3–1)
where x is a random number uniformly distributed in the interval [0,1]. For
example, to generate a vector y containing 1000 uniformly distributed random
numbers in the interval [2, 10], you type y = 8*rand(1,1000) + 2. You can
check the results with the mean, min, and max functions. You should obtain
values close to 6, 2, and 10, respectively.
You can use rand to generate random results for games involving dice, for
example, but you must use it to create integers. An easier way is to use the
randperm(n) function, which generates a random permutation of the integers
from 1 to n. For example, randperm(6) might generate the vector [3 2 6 4
1 5], or some other permutation of the numbers from 1 to 6. Note that
randperm calls rand and therefore changes the state of the generator.
With twister you control the internal state of the random number stream
used by rand and randn. Starting with MATLAB Version 7.7, the use of
twister is still supported for backwards compatibility but will be discontinued
eventually. For Version 7.7 and higher, you can generate the random number
stream using Randstream, which is an advanced topic. See the MATLAB
documentation.
Normally Distributed Random Numbers
In a sequence of normally distributed random numbers, the values near the mean
are more likely to occur. We have noted that the outcomes of many processes can
be described by the normal distribution. Although a uniformly distributed random variable has de nite upper and lower bounds, a normally distributed random variable does not.
The MATLAB function randn will generate a single number that is normally distributed with a mean equal to 0 and a standard deviation equal to 1.
Type randn(n) to obtain an n ⫻ n matrix of such numbers. Type
randn(m,n) to obtain an m ⫻ n matrix of random numbers.
The functions for retrieving and specifying the state of the normally distributed
random number generator are identical to those for the uniformly distributed generator, except that randn(...) replaces rand(...) in the syntax and ‘state’
is used instead of ‘twister’. These functions are summarized in Table 7.3–1.
You can generate a sequence of normally distributed numbers having a mean
␮ and standard deviation ␴ from a normally distributed sequence having a mean
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of 0 and a standard deviation of 1. You do this by multiplying the values by ␴ and
adding ␮ to each result. Thus if x is a random number with a mean of 0 and a
standard deviation of 1, use the following equation to generate a new random
number y having a standard deviation of ␴ and a mean of ␮.
y = ␴x + ␮
(7.3–2)
For example, to generate a vector y containing 2000 random numbers normally
distributed with a mean of 5 and a standard deviation of 3, you type y =
3*randn(1,2000) + 5. You can check the results with the mean and std
functions. You should obtain values close to 5 and 3, respectively.
Test Your Understanding
T7.3–1 Use MATLAB to generate a vector y containing 1800 random numbers
normally distributed with a mean of 7 and a standard deviation of 10.
Check your results with the mean and std functions. Why can’t you
use the min and max functions to check your results?
Functions of Random Variables
If y and x are linearly related as
y = bx + c
(7.3–3)
and if x is normally distributed with a mean ␮x and standard deviation ␴x , it can
be shown that the mean and standard deviation of y are given by
␮y = b␮x + c
(7.3–4)
␴y = |b|␴x
(7.3–5)
However, it is easy to see that the means and standard deviations do not combine
in a straightforward fashion when the variables are related by a nonlinear function. For example, if x is normally distributed with a mean of 0, and if y ⫽ x2, it
is easy to see that the mean of y is not 0, but is positive. In addition, y is not normally distributed.
Some advanced methods are available for deriving a formula for the mean
and variance of y ⫽ f(x), but for our purposes, the simplest way is to use random
number simulation.
It was noted in the previous section that the mean of the sum (or difference)
of two independent normally distributed random variables equals the sum (or difference) of their means, but the variance is always the sum of the two variances.
However, if z is a nonlinear function of x and y, then the mean and variance of z
cannot be found with a simple formula. In fact, the distribution of z will not even
be normal. This outcome is illustrated by the following example.
Statistical Analysis and Manufacturing Tolerances
Suppose you must cut a triangular piece off the corner of a square plate by measuring the
distances x and y from the corner (see Figure 7.3–1). The desired value of x is 10 in., and
EXAMPLE 7.3–1
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Plate
y
θ
x
Figure 7.3–1 Dimensions of a
triangular cut.
the desired value of ␪ is 20⬚. This requires that y ⫽ 3.64 in. We are told that measurements
of x and y are normally distributed with means of 10 and 3.64, respectively, with a
standard deviation equal to 0.05 in. Determine the standard deviation of ␪ and plot the
relative frequency histogram for ␪.
■ Solution
From Figure 7.3–1, we see that the angle ␪ is determined by ␪ ⫽ tan⫺1 (y/x). We can nd
the statistical distribution of ␪ by creating random variables x and y that have means of 10
and 3.64, respectively, with a standard deviation of 0.05. The random variable ␪ is then
found by calculating ␪ ⫽ tan⫺1 (y/x) for each random pair (x,y). The following script le
shows this procedure.
s = 0.05; % standard deviation of x and y
n = 8000; % number of random simulations
x = 10 + s*randn(1,n);
y = 3.64 + s*randn(1,n);
theta = (180/pi)*atan(y./x);
mean_theta = mean(theta)
sigma_theta = std(theta)
xp = 19:0.1:21;
z = hist(theta,xp);
yp = z/n;
bar(xp,yp),xlabel(‘Theta (degrees)’),...
ylabel(‘Relative Frequency’)
The choice of 8000 simulations was a compromise between accuracy and the amount of
time required to do the calculations. You should try different values of n and compare the
results. The results gave a mean of 19.9993⬚ for ␪ with a standard deviation of 0.2730⬚.
The histogram is shown in Figure 7.3–2. Although the plot resembles the normal distribution, the values of ␪ are not distributed normally. From the histogram we can calculate
that approximately 65 percent of the values of ␪ lie between 19.8 and 20.2. This range
corresponds to a standard deviation of 0.2⬚, not 0.273⬚ as calculated from the simulation
data. Thus the curve is not a normal distribution.
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0.14
Relative Frequency
0.12
0.1
0.08
0.06
0.04
0.02
0
18.5
19
19.5
20
Theta (degrees)
20.5
21
21.5
Figure 7.3–2 Scaled histogram of the angle ␪.
This example shows that the interaction of two of more normally distributed variables does not produce a result that is normally distributed. In general, the result is normally distributed if and only if the result is a linear combination of the variables.
7.4 Interpolation
Paired data might represent a cause and effect, or input-output relationship, such
as the current produced in a resistor as a result of an applied voltage, or a time
history, such as the temperature of an object as a function of time. Another type
of paired data represents a pro le, such as a road pro le (which shows the height
of the road along its length). In some applications we want to estimate a variable’s value between the data points. This process is called interpolation. In other
cases we might need to estimate the variable’s value outside the given data range.
This process is called extrapolation. Interpolation and extrapolation are greatly
aided by plotting the data. Such plots, some perhaps using logarithmic axes,
often help to discover a functional description of the data.
Suppose we have the following temperature measurements, taken once an
hour starting at 7:00 A.M. The measurements at 8 and 10 A.M. are missing for
some reason, perhaps because of equipment malfunction.
Time
Temperature (⬚F)
7 A.M.
9 A.M.
11 A.M.
12 noon
49
57
71
75
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Temperature Measurements at a Single Location
80
75
70
Temperature (deg F)
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60
55
50
45
7
7.5
8
8.5
9
9.5
Time (hr)
10
10.5
11
11.5
12
Figure 7.4–1 A plot of temperature data versus time.
A plot of these data is shown in Figure 7.4–1 with the data points connected by dashed lines. If we need to estimate the temperature at 10 A.M., we
can read the value from the dashed line that connects the data points at 9 and
11 A.M. From the plot we thus estimate the temperature at 8 A.M. to be 53⬚F
and at 10 A.M. to be 64⬚F. We have just performed linear interpolation on the
data to obtain an estimate of the missing data. Linear interpolation is so
named because it is equivalent to connecting the data points with a linear
function (a straight line).
Of course we have no reason to believe that the temperature follows the
straight lines shown in the plot, and our estimate of 64⬚F will most likely be
incorrect, but it might be close enough to be useful. Using straight lines to connect the data points is the simplest form of interpolation. Another function could
be used if we have a good reason to do so. Later in this section we use polynomial functions to do the interpolation.
Linear interpolation in MATLAB is obtained with the interp1 and
interp2 functions. Suppose that x is a vector containing the independent variable data and that y is a vector containing the dependent variable data. If x_int
is a vector containing the value or values of the independent variable at which we
wish to estimate the dependent variable, then typing interp1(x,y,x_int)
produces a vector the same size as x_int containing the interpolated values of
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y that correspond to x_int. For example, the following session produces an
estimate of the temperatures at 8 and 10 A.M. from the preceding data. The vectors x and y contain the times and temperatures, respectively.
>>x = [7, 9, 11, 12];
>>y = [49, 57, 71, 75];
>>x_int = [8, 10];
>>interp1(x,y,x_int)
ans =
53
64
You must keep in mind two restrictions when using the interp1 function.
The values of the independent variable in the vector x must be in ascending
order, and the values in the interpolation vector x_int must lie within the range
of the values in x. Thus we cannot use the interp1 function to estimate the
temperature at 6 A.M., for example.
The interp1 function can be used to interpolate in a table of values by
de ning y to be a matrix instead of a vector. For example, suppose that we now
have temperature measurements at three locations and the measurements at 8 and
10 A.M. are missing for all three locations. The data are as follows:
Temperature (ⴗF)
Time
Location 1
Location 2
Location 3
7 A.M.
9 A.M.
11 A.M.
12 noon
49
57
71
75
52
60
73
79
54
61
75
81
We de ne x as before, but now we de ne y to be a matrix whose three columns
contain the second, third, and fourth columns of the preceding table. The following session produces an estimate of the temperatures at 8 and 10 A.M. at each
location.
>>x = [7, 9, 11, 12]’;
>>y(:,1) = [49, 57, 71, 75]’;
>>y(:,2) = [52, 60, 73, 79]’;
>>y(:,3) = [54, 61, 75, 81]’;
>>x_int = [8, 10]’;
>>interp1(x,y,x_int)
ans =
53.0000
56.0000
57.5000
64.0000
65.5000
68.0000
Thus the estimated temperatures at 8 A.M. at each location are 53, 56, and 57.5⬚F,
respectively. At 10 A.M. the estimated temperatures are 64, 65.5, and 68⬚F. From
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this example we see that if the rst ar gument x in the interp1(x,y,x_int)
function is a vector and the second argument y is a matrix, then the function
interpolates between the rows of y and computes a matrix having the same
number of columns as y and the same number of rows as the number of values
in x_int.
Note that we need not de ne two separate vectors x and y. Rather, we can
de ne a single matrix that contains the entire table. For example, by de ning the
matrix temp to be the preceding table, the session will look like this:
>>temp(:,1) = [7, 9, 11, 12]’;
>>temp(:,2) = [49, 57, 71, 75]’;
>>temp(:,3) = [52, 60, 73, 79]’;
>>temp(:,4) = [54, 61, 75, 81]’;
>>x_int = [8, 10]’;
>>interp1(temp(:,1),temp(:,2:4),x_int)
ans =
53.0000
56.0000
57.5000
64.0000
65.5000
68.0000
Two-Dimensional Interpolation
Now suppose that we have temperature measurements at four locations at 7 A.M.
These locations are at the corners of a rectangle 1 mi wide and 2 mi long. Assigning a coordinate system origin (0,0) to the rst location, the coordinates of the
other locations are (1, 0), (1, 2), and (0, 2); see Figure 7.4–2. The temperature
measurements are shown in the gure. The temperature is a function of two
(0,0)
(1,0)
x
49°
54°
(0.6, 1.5)
53°
(0,2)
57°
(1,2)
y
Figure 7.4–2 Temperature
measurements at four locations.
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variables, the coordinates x and y. MATLAB provides the interp2 function
to interpolate functions of two variables. If the function is written as z ⫽ f(x,y)
and we wish to estimate the value of z for x ⫽ xi and y ⫽ yi, the syntax is
interp2(x,y,z,x_i,y_i).
Suppose we want to estimate the temperature at the point whose coordinates
are (0.6,1.5). Put the x coordinates in the vector x and the y coordinates in the
vector y. Then put the temperature measurements in a matrix z such that going
across a row represents an increase in x and going down a column represents an
increase in y. The session to do this is as follows:
>>x = [0,1];
>>y = [0,2];
>>z = [49,54;53,57]
z =
49
54
53
57
>>interp2(x,y,z,0.6,1.5)
ans =
54.5500
Thus the estimated temperature is 54.55⬚F.
The syntax of the interp1 and interp2 functions is summarized in
Table 7.4–1. MATLAB also provides the interpn function for interpolating
multidimensional arrays.
Cubic Spline Interpolation
High-order polynomials can exhibit undesired behavior between the data points,
and this can make them unsuitable for interpolation. A widely used alternative
procedure is to t the data points using a lower -order polynomial between each
pair of adjacent data points. This method is called spline interpolation and is so
named for the splines used by illustrators to draw a smooth curve through a set
of points.
Spline interpolation obtains an exact t that is also smooth. The most common procedure uses cubic polynomials, called cubic splines, and thus is called
Table 7.4–1 Linear interpolation functions
Command
Description
y_int=interp1(x,y,x_int)
Used to linearly interpolate a function of one
variable: y ⫽ f(x). Returns a linearly
interpolated vector y_int at the speci ed
value x_int, using data stored in x and y.
Used to linearly interpolate a function of two
variables: y ⫽ f(x, y). Returns a linearly
interpolated vector z_int at the speci ed
values x_int and y_int, using data stored
in x, y, and z.
z_int=interp2(x,y,z,x_,y_int)
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cubic spline interpolation. If the data are given as n pairs of (x, y) values, then
n ⫺ 1 cubic polynomials are used. Each has the form
yi(x) = ai(x - xi)3 + bi(x - xi)2 + ci(x - xi) + di
for xi ⱕ x ⱕ xi⫹1 and i ⫽ 1, 2, . . . , n ⫺ 1. The coef cients ai, bi, ci, and di for
each polynomial are determined so that the following three conditions are satised for each polynomial:
1. The polynomial must pass through the data points at its endpoints at xi and xi⫹1.
2. The slopes of adjacent polynomials must be equal at their common data point.
3. The curvatures of adjacent polynomials must be equal at their common
data point.
For example, a set of cubic splines for the temperature data given earlier follows
(y represents the temperature values, and x represents the hourly values). The data
are repeated here.
x
y
7
49
9
57
11
71
12
75
We will shortly see how to use MATLAB to obtain these polynomials. For 7 ⱕ
x ⱕ 9,
y1(x) = - 0.35(x - 7)3 + 2.85(x - 7)2 - 0.3(x - 7) + 49
For 9 ⱕ x ⱕ 11,
y2(x) = - 0.35(x - 9)3 + 0.75(x - 9)2 + 6.9(x - 9) + 57
For 11 ⱕ x ⱕ 12,
y3(x) = - 0.35(x - 11)3 - 1.35(x - 11)2 + 5.7(x - 11) + 71
MATLAB provides the spline command to obtain a cubic spline
interpolation. Its syntax is y_int = spline(x,y,x_int), where x and y
are vectors containing the data and x_int is a vector containing the values of
the independent variable x at which we wish to estimate the dependent variable y.
The result y_int is a vector the same size as x_int containing the interpolated values of y that correspond to x_int. The spline t can be plotted by
plotting the vectors x_int and y_int. For example, the following session
produces and plots a cubic spline t to the preceding data, using an increment of
0.01 in the x values.
>>x = [7,9,11,12];
>>y = [49,57,71,75];
>>x_int = 7:0.01:12;
>>y_int = spline(x,y,x_int);
>>plot(x,y,’o’,x,y,’— —‘,x_int,y_int),...
xlabel(‘Time (hr)’),ylabel(‘Temperature (deg F)’), ...
title(‘Measurements at a Single Location’), ...
axis([7 12 45 80])
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Measurements at a Single Location
80
75
Temperature (deg F)
70
65
60
55
50
45
7
7.5
8
8.5
9
9.5
Time (hr)
10
10.5
11
11.5
12
Figure 7.4–3 Linear and cubic-spline interpolation of temperature data.
The plot is shown in Figure 7.4–3. The dashed lines represent linear
interpolation, and the solid curve is the cubic spline. If we evaluate the spline
polynomial at x ⫽ 8, we obtain y(8) ⫽ 51.2⬚F. This estimate is different from the
53⬚F estimate obtained from linear interpolation. It is impossible to say which
estimate is more accurate without having greater understanding of the temperature dynamics.
We can obtain an estimate more quickly by using the following variation of
the interp1 function.
y_est = interp1(x,y,x_est,’spline’)
In this form the function returns a column vector y_est that contains the estimated values of y that correspond to the x values speci ed in the vector x_est,
using cubic spline interpolation.
In some applications it is helpful to know the polynomial coef cients, but
we cannot obtain the spline coef cients from the interp1 function. However,
we can use the form
[breaks, coeffs, m, n] = unmkpp(spline(x,y))
to obtain the coef cients of the cubic polynomials. The vector breaks
contains the x values of the data, and the matrix coeffs is an m ⫻ n matrix
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containing the coef cients of the polynomials. The scalars m and n give the
dimensions of the matrix coeffs; m is the number of polynomials, and n is
the number of coef cients for each polynomial (MATLAB will t a lower order polynomial if possible, so there can be fewer than four coef cients). For
example, using the same data, the following session produces the coef cients
of the polynomials given earlier:
>>x = [7,9,11,12];
>>y = [49,57,71,75];
>> [breaks, coeffs, m, n] = unmkpp(spline(x,y))
breaks =
7
9 11
12
coeffs =
-0.3500 2.8500 -0.3000 49.0000
-0.3500 0.7500 6.900 57.0000
-0.3500 -1.3500 5.7000 71.0000
m =
3
n =
4
The rst row of the matrix coeffs contains the coef cients of the rst polynomial, and so on. These functions are summarized in Table 7.4–2. The Basic
Table 7.4–2 Polynomial interpolation functions
Command
Description
y_est = interp1(x,y,x_est, method)
Returns a column vector y_est that contains the
estimated values of y that correspond to the x values
speci ed in the vector x_est, using interpolation
speci ed by method. The choices for method are
‘nearest’, ‘linear’, ‘spline’, ‘pchip’, and ‘cubic’.
Computes a cubic spline interpolation where x
and y are vectors containing the data and x_int
is a vector containing the values of the independent
variable x at which we wish to estimate the
dependent variable y. The result y_int is a
vector the same size as x_int containing the
interpolated values of y that correspond to x_int.
Similar to spline but uses piecewise cubic
Hermite polynomials for interpolation to preserve
shape and respect monotonicity.
Computes the coef cients of the cubic spline
polynomials for the data in x and y. The vector
breaks contains the x values, and the matrix
coeffs is an m ⫻ n matrix containing the
polynomial coef cients. The scalars m and n give
the dimensions of the matrix coeffs; m is the
number of polynomials, and n is the number of
coef cients for each polynomial.
y_int = spline(x,y,x_int)
y_int = pchip (x,y,x_int)
[breaks, coeffs, m, n] = unmkpp(spline(x,y))
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321
Fitting interface, which is available on the Tools menu of the Figure window, can
be used for cubic spline interpolation. See Section 6.3 for instructions for using
the interface.
As another example of interpolation, consider 10 evenly spaced data points
generated by the function y ⫽ 1/(3 ⫺ 3x ⫹ x2) over the range 0 ⱕ x ⱕ 4. The top
graph in Figure 7.4–4 shows the results of tting a cubic polynomial and an
eighth-order polynomial to the data. Clearly the cubic is not suitable for interpolation. As we increase the order of the tted polynomial, we nd that the polynomial does not pass through all the data points if the order is less than 7.
However, there are two problems with the eighth-order polynomial: we should
not use it to interpolate over the interval 0 ⬍ x ⬍ 0.5, and its coef cients must be
stored with very high accuracy if we use the polynomial to interpolate. The
1.4
data
cubic
eighth−order
1.2
1
y
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
x
2.5
3
3.5
4
1.4
data
spline
1.2
1
y
0.8
0.6
0.4
0.2
0
0
0.5
1
1.5
2
x
2.5
3
3.5
Figure 7.4–4 Top graph: Interpolation with a cubic polynomial and an eighth-order polynomial. Bottom graph:
Interpolation with a cubic spline.
4
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bottom graph in Figure 7.4–4 shows the results of tting a cubic spline, which is
clearly a better choice here.
Interpolation with Hermite Polynomials
The pchip function uses piecewise continuous Hermite interpolation polynomials (pchips). Its syntax is identical to that of the spline function. With pchip
the slopes at the data points are computed to preserve the “shape” of the data and
to “respect” monotonicity. That is, the tted function will be monotonic on intervals where the data are monotonic and will have a local extremum on intervals
where the data have a local extremum. The differences between the two functions
are that
■
■
■
The second derivatives are continuous with spline but may be discontinuous with pchip, so spline may give a smoother function.
Therefore, the spline function is more accurate if the data are “smoother.”
There are no overshoots and less oscillation in the function produced by
pchip, even if the data are not smooth.
Consider the data given by x ⫽ [0, 1, 2, 3, 4, 5] and y ⫽ [0, ⫺10, 60, 40, 41,
47]. The top graph in Figure 7.4–5 shows the results of tting a fth-order polynomial and a cubic spline to the data. Clearly the fth-order polynomial is less
suitable for interpolation because of the large excursions it makes, especially over
the ranges 0 ⬍ x ⬍ 1 and 4 ⬍ x ⬍ 5. These excursions are often seen with highorder polynomials. Here, the cubic spline is more useful. The bottom graph in
Figure 7.4–5 compares the results of a cubic spline t with a piecewise continuous Hermite polynomial t (using pchip), which is clearly a better choice here.
MATLAB provides a number of other functions to support interpolation for
three-dimensional data. See griddata, griddata3, griddatan, interp3,
and interpn in the MATLAB Help.
7.5 Summary
This chapter introduces MATLAB functions that have widespread and important
uses in statistics and data analysis. Section 7.1 gives an introduction to basic
statistics and probability, including histograms, which are specialized plots for
displaying statistical results. The normal distribution that forms the basis of
many statistical methods is covered in Section 7.2. Section 7.3 covers random
number generators and their use in simulation programs. Section 7.4 covers
interpolation methods, including linear and spline interpolation.
Now that you have nished this chapter , you should be able to use MATLAB to
■
■
■
Solve basic problems in statistics and probability.
Create simulations incorporating random processes.
Apply interpolation to data.
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323
80
60
40
y
20
0
data
fifth order
spline
−20
−40
−60
0
0.5
1
1.5
2
2.5
x
3
3.5
4
4.5
5
80
60
y
40
20
data
pchip
spline
0
−20
−40
0
0.5
1
1.5
2
2.5
x
3
3.5
4
4.5
5
Figure 7.4–5 Top graph: Interpolation with a fth-order polynomial and a cubic spline. Bottom graph: Interpolation
with piecewise continuous Hermite polynomials (pchip) and a cubic spline.
Key Terms with Page References
Absolute frequency, 297
Bins, 296
Cubic splines, 317
Error function, 305
Gaussian function, 303
Histogram, 296
Interpolation, 313
Mean, 296
Median, 296
Mode, 296
Normally distributed, 303
Normal or Gaussian function, 303
Relative frequency, 297
Scaled frequency histogram, 301
Standard deviation, 303
Uniformly distributed, 307
Variance, 303
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Problems
You can nd the answers to problems marked with an asterisk at the end of
the text.
Section 7.1
1.
The following list gives the measured gas mileage in miles per gallon for
22 cars of the same model. Plot the absolute frequency histogram and the
relative frequency histogram.
23
26
2.
26
24
25
22
27
25
25
26
24
24
22
24
23
24
25
27
26
23
Thirty pieces of structural timber of the same dimensions were subjected
to an increasing lateral force until they broke. The measured force in
pounds required to break them is given in the following list. Plot the absolute frequency histogram. Try bin widths of 50, 100, and 200 lb. Which
gives the most meaningful histogram? Try to nd a better value for the
bin width.
243
404
497
132
457
3.
25
24
236
464
500
196
347
389
605
535
217
628
137
577
660
143
123
441
569
417
372
231
865
205
439
675
725
The following list gives the measured breaking force in newtons for a
sample of 60 pieces of certain type of cord. Plot the absolute frequency
histogram. Try bin widths of 10, 30, and 50 N. Which gives the most
meaningful histogram? Try to nd a better value for the bin width.
311 138 340 199 270 255 332 279 231 296 198 269
257 236 313 281 288 225 216 250 259 323 280 205
279 159 276 354 278 221 192 281 204 361 321 282
254 273 334 172 240 327 261 282 208 213 299 318
356 269 355 232 275 234 267 240 331 222 370 226
Section 7.2
4.
For the data given in Problem 1:
a. Plot the scaled frequency histogram.
b. Compute the mean and standard deviation and use them to estimate the
lower and upper limits of gas mileage corresponding to 68 percent of
cars of this model. Compare these limits with those of the data.
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5.
For the data given in Problem 2:
a. Plot the scaled frequency histogram.
b. Compute the mean and standard deviation and use them to estimate the
lower and upper limits of strength corresponding to 68 and 96 percent
of such timber pieces. Compare these limits with those of the data.
6.
For the data given in Problem 3:
a. Plot the scaled frequency histogram.
b. Compute the mean and standard deviation, and use them to estimate
the lower and upper limits of breaking force corresponding to
68 and 96 percent of cord pieces of this type. Compare these limits
with those of the data.
7.* Data analysis of the breaking strength of a certain fabric shows that it is
normally distributed with a mean of 300 lb and a variance of 9.
a. Estimate the percentage of fabric samples that will have a breaking
strength no less than 294 lb.
b. Estimate the percentage of fabric samples that will have a breaking
strength no less than 297 lb and no greater than 303 lb.
8.
Data from service records show that the time to repair a certain machine
is normally distributed with a mean of 65 min and a standard deviation
of 5 min. Estimate how often it will take more than 75 min to repair a
machine.
9.
Measurements of a number of ttings show that the pitch diameter of the
thread is normally distributed with a mean of 8.007 mm and a standard
deviation of 0.005 mm. The design speci cations require that the pitch
diameter be 8 ⫾ 0.01 mm. Estimate the percentage of ttings that will be
within tolerance.
10.
A certain product requires that a shaft be inserted into a bearing.
Measurements show that the diameter d1 of the cylindrical hole in the
bearing is normally distributed with a mean of 3 cm and a variance of
0.0064. The diameter d2 of the shaft is normally distributed with a mean
of 2.96 cm and a variance of 0.0036.
a. Compute the mean and the variance of the clearance c ⫽ d1 ⫺ d2.
b. Find the probability that a given shaft will not t into the bearing.
(Hint: Find the probability that the clearance is negative.)
11.* A shipping pallet holds 10 boxes. Each box holds 300 parts of different
types. The part weight is normally distributed with a mean of 1 lb and a
standard deviation of 0.2 lb.
a. Compute the mean and standard deviation of the pallet weight.
b. Compute the probability that the pallet weight will exceed 3015 lb.
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12.
A certain product is assembled by placing three components end to end.
The components’ lengths are L1, L2, and L3. Each component is manufactured on a different machine, so the random variations in their lengths are
independent of one another. The lengths are normally distributed with
means of 1, 2, and 1.5 ft and variances of 0.00014, 0.0002, and 0.0003,
respectively.
a. Compute the mean and variance of the length of the assembled
product.
b. Estimate the percentage of assembled products that will be no less than
4.48 and no more than 4.52 ft in length.
Section 7.3
13.
Use a random number generator to produce 1000 uniformly distributed
numbers with a mean of 10, a minimum of 2, and a maximum of 18.
Obtain the mean and the histogram of these numbers, and discuss
whether they appear uniformly distributed with the desired mean.
14.
Use a random number generator to produce 1000 normally distributed
numbers with a mean of 20 and a variance of 4. Obtain the mean,
variance, and histogram of these numbers, and discuss whether they
appear normally distributed with the desired mean and variance.
15.
The mean of the sum (or difference) of two independent random variables
equals the sum (or difference) of their means, but the variance is always the
sum of the two variances. Use random number generation to verify this
statement for the case where z ⫽ x ⫹ y, where x and y are independent and
normally distributed random variables. The mean and variance of x are
␮x ⫽ 8 and ␴2x ⫽ 2. The mean and variance of y are ␮y ⫽ 15 and
␴ 2y ⫽ 4. Find the mean and variance of z by simulation, and compare
the results with the theoretical prediction. Do this for 100, 1000, and
5000 trials.
16.
Suppose that z ⫽ xy, where x and y are independent and normally
distributed random variables. The mean and variance of x are ␮x ⫽ 10 and
␴ 2x ⫽ 2. The mean and variance of y are ␮y ⫽ 15 and ␴2y ⫽ 3. Find the
mean and variance of z by simulation. Does ␮z ⫽ ␮x ␮y? Does ␴ 2z ⫽ ␴ 2x ␴ 2y?
Do this for 100, 1000, and 5000 trials.
17.
Suppose that y ⫽ x2, where x is a normally distributed random variable
with a mean and variance of ␮x ⫽ 0 and ␴ 2x ⫽ 4. Find the mean and
variance of y by simulation. Does ␮y ⫽ ␮2x? Does ␴y ⫽ ␴2x ? Do this for
100, 1000, and 5000 trials.
18.* Suppose you have analyzed the price behavior of a certain stock by
plotting the scaled frequency histogram of the price over a number of
months. Suppose that the histogram indicates that the price is normally
distributed with a mean $100 and a standard deviation of $5. Write a
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Problems
MATLAB program to simulate the effects of buying 50 shares of this
stock whenever the price is below the $100 mean, and selling all your
shares whenever the price is above $105. Analyze the outcome of this
strategy over 250 days (the approximate number of business days in a
year). Define the profit as the yearly income from selling stock plus
the value of the stocks you own at year’s end, minus the yearly cost
of buying stock. Compute the mean yearly profit you would expect
to make, the minimum expected yearly profit, the maximum expected
yearly profit, and the standard deviation of the yearly profit. The
broker charges 6 cents per share bought or sold with a minimum fee
of $40 per transaction. Assume you make only one transaction
per day.
19.
Suppose that data show that a certain stock price is normally distributed
with a mean of $150 and a variance of 100. Create a simulation to compare the results of the following two strategies over 250 days. You start
the year with 1000 shares. With the rst strategy , every day the price is
below $140 you buy 100 shares, and every day the price is above
$160 you sell all the shares you own. With the second strategy, every
day the price is below $150 you buy 100 shares, and every day the price
is above $160 you sell all the shares you own. The broker charges
5 cents per share traded with a minimum of $35 per transaction.
20.
Write a script le to simulate 100 plays of a game in which you ip two
coins. You win the game if you get two heads, lose if you get two tails, and
ip again if you get one head and one tail. Create three user -de ned functions to use in the script. Function ip simulates the ip of one coin, with
the state s of the random number generator as the input argument, and the
new state s and the result of the ip (0 for a tail and 1 for a head) as the
outputs. Function ips simulates the ipping of two coins and calls ip.
The input of ips is the state s, and the outputs are the new state s and
the result (0 for two tails, 1 for a head and a tail, and 2 for two heads).
Function match simulates a turn at the game. Its input is the state s, and
its outputs are the result (1 for win, 0 for lose) and the new state s. The
script should reset the random number generator to its initial state, compute
the state s, and pass this state to the user-de ned functions.
21.
Write a script le to play a simple number guessing game as follows.
The script should generate a random integer in the range 1, 2, 3, . . . , 14,
15. It should provide for the player to make repeated guesses of the
number, and it should indicate if the player has won or give the player a
hint after each wrong guess. The responses and hints are as follows:
■
■
■
■
“You won” and then stop the game.
“Very close” if the guess is within 1 of the correct number.
“Getting close” if the guess is within 2 or 3 of the correct number.
“Not close” if the guess is not within 3 of the correct number.
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CHAPTER 7 Statistics, Probability, and Interpolation
Section 7.4
22.* Interpolation is useful when one or more data points are missing. This situation often occurs with environmental measurements, such as temperature, because of the dif culty of making measurements around the clock.
The following table of temperature versus time data is missing readings at
5 and 9 hours. Use linear interpolation with MATLAB to estimate the
temperature at those times.
Time (hours, P.M.)
1
2
3
4
5
6
7
8
9
10
11
12
Temperature (⬚C)
10
9
18
24
?
21
20
18
?
15
13
11
23.
The following table gives temperature data in ⬚C as a function of time of
day and day of the week at a speci c location. Data are missing for the
entries marked with a question mark (?). Use linear interpolation with
MATLAB to estimate the temperature at the missing points.
Day
24.
Hour
Mon
Tues
Wed
Thurs
Fri
1
2
3
4
5
17
13
14
17
23
15
?
14
15
18
12
8
9
14
17
16
11
?
15
20
16
12
15
19
24
Computer-controlled machines are used to cut and to form metal and
other materials when manufacturing products. These machines often use
cubic splines to specify the path to be cut or the contour of the part to be
shaped. The following coodinates specify the shape of a certain car’s
front fender. Fit a series of cubic splines to the coordinates, and plot the
splines along with the coordinate points.
x (ft)
0
0.25
0.75
1.25
1.5
1.75
1.875
2
2.125
2.25
y (ft)
1.2
1.18
1.1
1
0.92
0.8
0.7
0.55
0.35
0
25.
The following data are the measured temperature T of water owing from
a hot water faucet after it is turned on at time t ⫽ 0.
t (sec) T (ⴗF)
t (sec)
T (ⴗF)
0
1
2
3
4
5
6
7
8
9
10
109.3
110.2
110.5
109.9
110.2
72.5
78.1
86.4
92.3
110.6
111.5
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Problems
a. Plot the data, connecting them rst with straight lines and then with a
cubic spline.
b. Estimate the temperature values at the following times, using linear interpolation and then cubic spline interpolation: t ⫽ 0.6, 2.5, 4.7, 8.9.
c. Use both the linear and cubic spline interpolations to estimate the time
it will take for the temperature to equal the following values: T ⫽ 75,
85, 90, 105.
329
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© Stocktrek/age fotostock/RF
Engineering in the
21st Century. . .
International Engineering
T
he end of the Cold War has resulted in increased international cooperation. A vivid example of this is the cooperation between the Russian and
American space agencies. In the rst joint mission the U.S. space shuttle
Atlantis docked with the Russian space station Mir. Now Russia, the United
States, and several other countries are collaborating on a more ambitious space
project: the International Space Station (ISS). The launch of the rst element of
the station, a Russian-built cargo block, occurred in 1998. The rst permanent
crew arrived in 2000. The station will have a crew of seven, will weigh over
1 000 000 lb, and will be 290 ft long and 356 ft wide. To assemble it will require
28 U.S. shuttle ights and 41 Russian ights over a period of 10 years.
The new era of international cooperation has resulted in increased integration
of national economies into one global economy. For example, many components
for cars assembled in the United States are supplied by companies in other countries. In the 21st century, improved communications and computer networks will
enable engineering teams in different countries to work simultaneously on different aspects of the product’s design, employing a common design database. To
prepare for such a future in international engineering, students need to become
familiar with other cultures and languages as well as pro cient in computer -aided
design.
MATLAB is available in international editions and has been widely used in
Europe for many years. Thus engineers familiar with MATLAB will already
have a head start in preparing for a career in international engineering. ■
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C H A P T E R
8
Linear Algebraic
Equations
OUTLINE
8.1 Matrix Methods for Linear Equations
8.2 The Left Division Method
8.3 Underdetermined Systems
8.4 Overdetermined Systems
8.5 A General Solution Program
8.6 Summary
Problems
Linear algebraic equations such as
5x - 2y = 13
7x + 3y = 24
occur in many engineering applications. For example, electrical engineers use
them to predict the power requirements for circuits; civil, mechanical, and
aerospace engineers use them to design structures and machines; chemical engineers use them to compute material balances in chemical processes; and industrial engineers apply them to design schedules and operations. The examples and
homework problems in this chapter explore some of these applications.
Linear algebraic equations can be solved “by hand” using pencil and paper,
by calculator, or with software such as MATLAB. The choice depends on the
circumstances. For equations with only two unknown variables, hand solution is
easy and adequate. Some calculators can solve equation sets that have many variables. However, the greatest power and exibility is obtained by using software.
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CHAPTER 8 Linear Algebraic Equations
For example, MATLAB can obtain and plot equation solutions as we vary one or
more parameters.
Systematic solution methods have been developed for sets of linear equations.
In Section 8.1 we introduce some matrix notation that is required for use with
MATLAB and that is also useful for expressing solution methods in a compact
way. The conditions for the existence and uniqueness of solutions are then introduced. Methods using MATLAB are treated in four sections: Section 8.2 covers
the left division method for solving equation sets that have unique solutions.
Section 8.3 covers the case where the equation set does not contain enough information to determine all the unknown variables. This is the underdetermined case.
The overdetermined case occurs when the equation set has more independent
equations than unknowns (Section 8.4). A general solution program is given in
Section 8.5.
8.1 Matrix Methods for Linear Equations
Sets of linear algebraic equations can be expressed as a single equation, using matrix notation. This standard and compact form is useful for expressing solutions
and for developing software applications with an arbitrary number of variables. In
this application, a vector is taken to be a column vector unless otherwise speci ed.
Matrix notation enables us to represent multiple equations as a single matrix
equation. For example, consider the following set.
2x1 + 9x2 5
3x1 4x2 7
This set can be expressed in vector-matrix form as
c
2
3
9 x1
5
d c d c d
- 4 x2
7
which can be represented in the following compact form
Ax = b
(8.1–1)
where we have de ned the following matrices and vectors:
A c
2
3
9
d
-4
x c
x1
d
x2
5
b c d
7
In general, the set of m equations in n unknowns can be expressed in the form
Equation (8.1–1), where A is m n, x is n 1, and b is m 1.
Matrix Inverse
The solution of the scalar equation ax b is x b/a if a 0. The division operation of scalar algebra has an analogous operation in matrix algebra. For example, to solve the matrix equation (8.1–1) for x, we must somehow “divide” b by A.
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8.1
Matrix Methods for Linear Equations
333
The procedure for doing this is developed from the concept of a matrix inverse.
The inverse of a matrix A is denoted by A⫺1 and has the property that
A⫺1A ⫽ AA⫺1 ⫽ I
where I is the identity matrix. Using this property, we multiply both sides of Equation (8.1–1) from the left by A⫺1 to obtain A⫺1Ax ⫽ A⫺1b. Because A⫺1Ax ⫽ Ix ⫽ x,
we obtain the solution
x ⫽ A⫺1b
(8.1–2)
The inverse of a matrix A is de ned only if A is square and nonsingular. A matrix
is singular if its determinant |A| is zero. If A is singular, then a unique solution to
Equation (8.1–1) does not exist. The MATLAB functions inv(A) and det(A)
compute the inverse and the determinant of the matrix A. If the inv(A) function
is applied to a singular matrix, MATLAB will issue a warning to that effect.
An ill-conditioned set of equations is a set that is close to being singular. The
ill-conditioned status depends on the accuracy with which the solution calculations are made. When internal numerical accuracy used by MATLAB is insuf cient to obtain a solution, it prints the message warning that the matrix is close to
singular and that the results might be inaccurate.
For a 2 ⫻ 2 matrix A,
A ⫽ c
a b
d
c d
A⫺1 ⫽
1
d
c
ad ⫺ bc ⫺c
SINGULAR
MATRIX
ILL-CONDITIONED
SET
⫺b
d
a
where det(A) ⫽ ad ⫺ bc. Thus A is singular if ad ⫺ bc ⫽ 0.
The Matrix Inverse Method
EXAMPLE 8.1–1
Solve the following equations, using the matrix inverse.
2x1 + 9x2 ⫽ 5
3x1 ⫺ 4x2 ⫽ 7
■ Solution
The matrix A and the vector b are
A ⫽ c
The session is
2
3
9
d
⫺4
5
b ⫽ c d
7
>>A = [2,9;3,-4]; b = [5;7];
>>x = inv(A)*b
x =
2.3714
0.0286
The solution is x1 ⫽ 2.3714 and x2 ⫽ 0.0286. MATLAB did not issue a warning, so the
solution is unique.
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CHAPTER 8 Linear Algebraic Equations
The solution form x ⴝ A1b is rarely applied in practice to obtain numerical
solutions to sets of many equations, because calculation of the matrix inverse is
likely to introduce greater numerical inaccuracy than the left division method to be
introduced.
Test Your Understanding
T8.1–1 For what values of c will the following set (a) have a unique solution and
(b) have an in nite number of solutions? Find the relation between x1
and x2 for these solutions.
6x1 + cx2 0
2x1 + 4x2 0
(Answers: (a) c 12, x1 x2 0; (b) c 12, x1 2x2)
T8.1–2 Use the matrix inverse method to solve the following set.
3x1 4x2 5
6x1 10x2 2
(Answer: x1 7, x2 4)
T8.1–3 Use the matrix inverse method to solve the following set.
3x1 4x2 5
6x1 8x2 2
(Answer: No solution.)
Existence and Uniqueness of Solutions
MATRIX RANK
The matrix inverse method will warn us if a unique solution does not exist, but
it does not tell us whether there is no solution or an in nite number of solutions. In addition, the method is limited to cases where the matrix A is square,
that is, cases where the number of equations equals the number of unknowns.
For this reason we now introduce a method that allows us to determine easily
whether an equation set has a solution and whether it is unique. The method requires the concept of the rank of a matrix.
Consider the 3 3 determinant
3
|A| = † 6
9
-4 1
10 2 † = 0
-7 3
(8.1–3)
If we eliminate one row and one column in the determinant, we are left with a
2 2 determinant. Depending on which row and column we choose to eliminate, there are nine possible 2 2 determinants we can obtain. These are
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8.2
The Left Division Method
called subdeterminants. For example, if we eliminate the second row and third
column, we obtain
`
3
9
4
` 3( 7) 9( 4) 15
7
335
SUBDETERMINANT
Subdeterminants are used to de ne the rank of a matrix. The de nition of
matrix rank is as follows.
De nition of Matrix Rank. An m n matrix A has a rank r 1 if and only if
|A| contains a nonzero r r determinant and every square subdeterminant with
r 1 or more rows is zero.
For example, the rank of A in Equation (8.1–3) is 2 because |A| 0 while |A|
contains at least one nonzero 2 2 subdeterminant. To determine the rank of a
matrix A in MATLAB, type rank(A). If A is n n, its rank is n if det(A) 0.
We can use the following test to determine if a solution exists to Ax b
and whether it is unique. The test requires that we rst form the augmented
matrix [A b].
AUGMENTED
MATRIX
Existence and Uniqueness of Solutions. The set Ax b with m equations and
n unknowns has solutions if and only if (1) rank(A) rank([A b]). Let r rank(A). If condition (1) is satis ed and if r n, then the solution is unique. If
condition (1) is satis ed but r < n, there are an in nite number of solutions, and
r unknown variables can be expressed as linear combinations of the other n r
unknown variables, whose values are arbitrary.
Homogeneous case. The homogeneous set Ax 0 is a special case in which
b 0. For this case, rank(A) rank([A b]) always, and thus the set always
has the trivial solution x 0. A nonzero solution, in which at least one
unknown is nonzero, exists if and only if rank(A) < n. If m < n, the homogeneous set always has a nonzero solution.
This test implies that if A is square and of dimension n n, then
rank([A b]) rank(A), and a unique solution exists for any b if rank(A) n.
8.2 The Left Division Method
MATLAB provides the left division method for solving the equation set Ax b.
This method is based on Gauss elimination. To use the left division method to
solve for x, you type x = A\b. If |A| 0 or if the number of equations does not
equal the number of unknowns, then you need to use the other methods to be
presented later.
Left Division Method with Three Unknowns
EXAMPLE 8.2–1
Use the left division method to solve the following set.
3x1 + 2x2 9x3 65
9x1 5x2 + 2x3 16
6x1 + 7x2 + 3x3 5
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CHAPTER 8 Linear Algebraic Equations
■ Solution
The matrices A and b are
A The session is
3
9
J
6
9
2
K
3
2
5
7
b 65
16
J
K
5
>>A = [3,2,-9;-9,-5,2;6,7,3];
>>rank(A)
ans =
3
Because A is 3 3 and rank(A) 3, which is the number of unknowns, a unique solution
exists. It is obtained by continuing the session as follows.
>>b = [-65;16;5];
>>x = A\b
x =
2.0000
-4.0000
7.0000
This answer gives the vector x, which corresponds to the solution x1 2, x2 4, x3 7.
For the solution x A1b, vector x is proportional to the vector b. We can
use this linearity property to obtain a more generally useful algebraic solution in
cases where the right-hand sides are all multiplied by the same scalar. For example, suppose the matrix equation is Ay bc, where c is a scalar. The solution
is y A1bc xc. Thus if we obtain the solution to Ax b, the solution to
Ay bc is given by y xc.
Calculation of Cable Tension
EXAMPLE 8.2–2
A mass m is suspended by three cables attached at three points B, C, and D, as shown in
Figure 8.2–1. Let T1, T2, and T3 be the tensions in the three cables AB, AC, and AD, respectively. If the mass m is stationary, the sum of the tension components in the x, in the y, and
in the z directions must each be zero. This gives the following three equations:
T1
135
3T2
134
3T1
135
5T2
5T1
135
+
T3
134
+
+
142
4T3
142
5T3
142
0
0
mg 0
Determine T1, T2, and T3 in terms of an unspeci ed value of the weight mg.
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8.2
1m
The Left Division Method
3m
B
3m
4m
C
D
5m
z
y
1m
x
A
m
Figure 8.2–1 A mass suspended by three cables.
■ Solution
If we set mg 1, the equations have the form AT b where
1
135
A ≥135
3
5
135
3
134
0
5
134
1
142
1 4 ¥
The script le to solve this system is
42
5
142
T T1
T2
J K
T3
b 0
0
J K
1
% File cable.m
s34 = sqrt(34); s35 = sqrt(35); s42 = sqrt(42);
A1 = [1/s35, -3/s34, 1/s42];
A2 = [3/s35, 0, -4/s42];
A3 = [5/s35, 5/s34, 5/s42];
A = [A1; A2; A3];
b = [0; 0; 1];
rank(A)
rank([A, b])
T = A\b
When this le is executed by typing cable, we nd that rank( A) rank ([A b]) 3 and
obtain the values T1 0.5071, T2 0.2915, and T3 0.4166. Because A is 3 3 and
rank(A) 3, which is the number of unknowns, the solution is unique. Using the linearity
property, we multiply these results by mg and obtain the general solution T1 0.5071 mg,
T2 0.2915 mg, and T3 0.4166 mg.
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CHAPTER 8 Linear Algebraic Equations
Linear equations are useful in many engineering elds. Electric circuits are
a common source of linear equation models. The circuit designer must be able to
solve them to predict the currents that will exist in the circuit. This information
is often needed to determine the power supply requirements among other things.
An Electric Resistance Network
EXAMPLE 8.2–3
The circuit shown in Figure 8.2–2 has ve resistances and two applied voltages.
Assuming that the positive directions of current ow are in the directions shown in the
gure, Kirchhof f’s voltage law applied to each loop in the circuit gives
␷1 + R1i1 + R4i4 0
R4i4 + R2i2 + R5i5 0
R5i5 + R3i3 + ␷2 0
Conservation of charge applied at each node in the circuit gives
i1 i2 + i4
i2 i3 + i5
You can use these two equations to eliminate i4 and i5 from the rst three equations. The
result is
(R1 + R4)i1 R4i2 ␷1
R4i1 + (R2 + R4 + R5)i2 R5i3 0
R5i2 (R3 + R5)i3 ␷2
Thus we have three equations in the three unknowns i1, i2, and i3.
Write a MATLAB script le that uses given values of the applied voltages 1 and 2
and given values of the ve resistances to solve for the currents i1, i2, and i3. Use the
R1
i1
i2
R2
i3
i4
R3
i5
+
+
v1
R4
R5
–
Figure 8.2–2 An electric resistance network.
v2
–
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The Left Division Method
339
program to nd the currents for the case R1 5, R2 100, R3 200, R4 150, and R5 250 k and ␷1 100 and ␷ 2 50 V. (Note that 1 k 1000 .)
■ Solution
Because there are as many unknowns as equations, there will be a unique solution if
A 0; in addition, the left division method will generate an error message if A = 0.
The following script le, named resist.m, uses the left division method to solve the
three equations for i1, i2, and i3.
% File resist.m
% Solves for the currents i_1, i_2, i_3
R = [5,100,200,150,250]*1000;
v1 = 100; v2 = 50;
A1 = [R(1) + R(4), -R(4), 0];
A2 = [-R(4), R(2) + R(4) + R(5), -R(5)];
A3 = [0, R(5), -(R(3) + R(5))];
A = [A1; A2; A3];
b=[v1; 0; v2];
current = A\b;
disp(‘The currents are:’)
disp(current)
The row vectors A1, A2, and A3 were de ned to avoid typing the lengthy expression for
A in one line. This script is executed from the command prompt as follows:
>>resist
The currents are:
1.0e-003*
0.9544
0.3195
0.0664
Because MATLAB did not generate an error message, the solution is unique. The
currents are i1 0.9544, i2 0.3195, and i3 0.0664 mA, where 1 mA 1 milliampere 0.001 A.
Ethanol Production
Engineers in the food and chemical industries use fermentation in many processes. The
following equation describes Baker’s yeast fermentation.
a(C6H12O6) + b(O2) + c(NH3)
: C6H10NO3 + d(H2O) + e(CO2) + f (C2H6O)
The variables a, b, . . . , f represent the masses of the products involved in the reaction.
In this formula C6H12O6 represents glucose, C6H10NO3 represents yeast, and C2H6O
represents ethanol. This reaction produces ethanol in addition to water and carbon
EXAMPLE 8.2–4
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CHAPTER 8 Linear Algebraic Equations
dioxide. We want to determine the amount of ethanol f produced. The number of C, O,
N, and H atoms on the left must balance those on the right side of the equation. This
gives four equations:
6a = 6 + e + 2f
6a + 2b = 3 + d + 2e + f
c = 1
12a + 3c = 10 + 2d + 6f
The fermenter is equipped with an oxygen sensor and a carbon dioxide sensor. These
enable us to compute the respiratory quotient R:
R=
CO2
e
=
O2
b
Thus the fth equation is Rb - e = 0. The yeast yield Y (grams of yeast produced per
gram of glucose consumed) is related to a as follows
Y =
144
180a
where 144 is the molecular weight of yeast and 180 is the molecular weight of glucose.
By measuring the yeast yield Y we can compute a as follows: a = 144>180Y. This is the
sixth equation.
Write a user-de ned function that computes f, the amount of ethanol produced, with
R and Y as the function’s arguments. Test your function for two cases where Y is measured
to be 0.5: (a) R 1.1 and (b) R 1.05.
■ Solution
First note that there are only four unknowns because the third equation directly gives
c 1, and the sixth equation directly gives a = 144>180Y. To write these equations in
matrix form, let x1 = b, x2 = d, x3 = e, and x4 = f. Then the equations can be written as
-x3 - 2x4 = 6 - 6(144>180Y)
2x1 - x2 - 2x3 - x4 = 3 - 6(144>180Y)
-2x2 - 6x4 = 7 - 12(144>180Y)
Rx1 - x3 = 0
In matrix form these become
0
2
≥
0
R
0
-1
-2
0
-1
-2
0
-1
-2
x1
6 - 6(144>180Y)
-1
x
3 - 6(144>180Y)
¥ ≥ 2¥ = ≥
¥
-6
x3
7 - 12(144>180Y)
0
x4
0
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The function le is shown below .
function E = ethanol(R,Y)
% Computes ethanol produced from yeast reaction.
A = [0,0,-1,-2;2,-1,-2,-1;...
0,-2,0,-6;R,0,-1,0];
b = [6-6*(144./(180*Y));3-6*(144./(180*Y));...
7-12*(144./(180*Y));0];
x = A\b;
E = x(4);
The session is as follows:
>>ethanol(1.1,0.5)
ans =
0.0654
>>ethanol(1.05,0.5)
ans =
-0.0717
The negative value for E in the second case indicates that ethanol is being consumed rather
than produced.
Test Your Understanding
T8.2–1 Use the left division method to solve the following set.
5x1 - 3x2 = 21
7x1 - 2x2 = 36
(Answers: x1 6, x2 3)
8.3 Underdetermined Systems
An underdetermined system does not contain enough information to determine
all the unknown variables, usually but not always because it has fewer equations
than unknowns. Thus an in nite number of solutions can exist, with one or more
of the unknowns dependent on the remaining unknowns. The left division
method works for square and nonsquare A matrices. However, if A is not square,
the left division method can give answers that might be misinterpreted. We will
show how to interpret MATLAB results correctly.
When there are fewer equations than unknowns, the left division method might
give a solution with some of the unknowns set equal to zero, but this is not the general solution. An in nite number of solutions might exist even when the number of
equations equals the number of unknowns. This can occur when |A| 0. For such
systems the left division method generates an error message warning us that the
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PSEUDOINVERSE
METHOD
MINIMUM NORM
SOLUTION
EXAMPLE 8.3–1
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matrix A is singular. In such cases the pseudoinverse method x = pinv(A)*b
gives one solution, the minimum norm solution. In cases where there are an in nite
number of solutions, the rref function can be used to express some of the unknowns in terms of the remaining unknowns, whose values are arbitrary.
An equation set can be underdetermined even though it has as many equations as unknowns. This can happen if some of the equations are not independent.
Determining by hand whether all the equations are independent might not be easy,
especially if the set has many equations, but it is easily done in MATLAB.
An Underdetermined Set with Three Equations and Three Unknowns
Show that the following set does not have a unique solution. How many of the unknowns
will be undetermined? Interpret the results given by the left division method.
2x1 - 4x2 + 5x3 = - 4
- 4x1 - 2x2 + 3x3 = 4
2x1 + 6x2 - 8x3 = 0
■ Solution
A MATLAB session to check the ranks is
>>A = [2,-4,5;-4,-2,3;2,6,-8];
>>b = [-4;4;0];
>>rank(A)
ans =
2
>>rank([A, b])
ans =
2
>>x = A\b
Warning: Matrix is singular to working precision.
ans =
NaN
NaN
NaN
Because the ranks of A and [A b] are equal, a solution exists. However, because the number of unknowns is 3 and is 1 greater than the rank of A, one of the unknowns will be undetermined. An in nite number of solutions exist, and we can solve for only two of the
unknowns in terms of the third unknown. The set is underdetermined because there are
fewer than three independent equations; the third equation can be obtained from the rst
two. To see this, add the rst and second equations, to obtain 2x1 6x2 8x3 0,
which is equivalent to the third equation.
Note that we could also tell that the matrix A is singular because its rank is less than 3.
If we use the left division method, MATLAB returns a message warning that the problem
is singular, and it does not produce an answer.
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343
The pinv Function and the Euclidean Norm
The pinv function (which stands for “pseudoinverse”) can be used to obtain a
solution of an underdetermined set. To solve the equation set Ax b using the
pinv function, you type x = pinv(A)*b. The pinv function gives a solution
that gives the minimum value of the Euclidean norm, which is the magnitude of
the solution vector x. The magnitude of a vector v in three-dimensional space,
having components x, y, z, is 2x2 + y2 + z2. It can be computed using matrix
multiplication and the transpose as follows.
2v v =
T
x
[x y z] y = 2x2 + y2 + z2
JzK
T
The generalization of this formula to an n-dimensional vector v gives the magnitude of the vector and is the Euclidean norm N. Thus
N = 2vTv
(8.3–1)
The MATLAB function norm(v) computes the Euclidean norm.
A Statically Indeterminate Problem
EXAMPLE 8.3–2
Determine the forces in the three equally spaced supports that hold up a light xture. The
supports are 5 ft apart. The xture weighs 400 lb, and its mass center is 4 ft from the right
end. Obtain the solution using the MATLAB left-division method and the pseudoinverse
method.
■ Solution
Figure 8.3–1 shows the xture and the free-body diagram, where T1, T2, and T3 are the
tension forces in the supports. For the xture to be in equilibrium, the vertical forces must
cancel, and the total moments about an arbitrary xed point—say , the right endpoint—
must be zero. These conditions give the two equations
T1 + T2 + T3 - 400 = 0
400(4) - 10T1 - 5T2 = 0
or
T1 + T2 + T3 = 400
(8.3–2)
10T1 + 5T2 + 0T3 = 1600
(8.3–3)
Because there are more unknowns than equations, the set is underdetermined. Thus we
cannot determine a unique set of values for the forces. Such a problem, when the equations of statics do not give enough equations, is called statically indeterminate. These
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5 ft
5 ft
400 lb
(a)
T1
4 ft
T2
(b)
T3
400
Figure 8.3–1 A light xture and its free-body diagram.
equations can be written in the matrix form AT b as follows:
c
1
10
The MATLAB session is
>>A = [1,1,1;10,5,0];
>>b = [400;1600];
>>rank(A)
ans =
2
>>rank([A, b])
ans =
2
>>T = A\b
T =
160.0000
0
240.0000
1
5
T1
1
400
d T2 = c
d
0 J K
1600
T3
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>>T = pinv(A)*b
T =
93.3333
133.3333
173.3333
The left division answer corresponds to T1 160, T2 0, and T3 240. This illustrates
how the MATLAB left division operator produces a solution with one or more variables
set to zero, for underdetermined sets having more unknowns than equations.
Because the ranks of A and [A b] are both 2, a solution exists, but it is not unique. Because the number of unknowns is 3, and is 1 greater than the rank of A, an in nite number of
solutions exist, and we can solve for only two of the unknowns in terms of the third.
The pseudoinverse solution gives T1 93.3333, T2 133.3333, and T3 173.3333.
This is the minimum norm solution for real values of the variables. The minimum norm
solution consists of the real values of T1, T2, and T3 that minimize
N = 2T 21 + T 22 + T 23
To understand what MATLAB is doing, note that we can solve Equations (8.3–2) and
(8.3–3) to obtain T1 and T2 in terms of T3 as T1 T3 80 and T2 480 2T3. Then the
Euclidean norm can be expressed as
N = 2(T3 - 80)2 + (480 - 2T3)2 + T23 = 26T23 - 2080T3 + 236,800
The real value of T3 that minimizes N can be found by plotting N versus T3, or by using
calculus. The answer is T3 173.3333, the same as the minimum norm solution given by
the pseudoinverse method.
Where there are an in nite number of solutions, we must decide whether the
solutions given by the left division and the pseudoinverse methods are useful for
applications. This must be done in the context of the speci c application.
Test Your Understanding
T8.3–1 Find two solutions to the following set.
x1 + 3x2 + 2x3 = 2
x1 + x2 + x3 = 4
(Answer: Minimum norm solution: x1 4.33, x2 1.67, x3 1.34.
Left division solution: x1 5, x2 1, x3 0.)
The Reduced Row Echelon Form
We can express some of the unknowns in an underdetermined set as functions
of the remaining unknowns. In Example 8.3–2, we wrote the solutions for two of
the unknowns in terms of the third: T1 T3 80 and T2 480 2T3. These two
equations are equivalent to
T1 - T3 = - 80
T2 + 2T3 = 480
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In matrix form these are
c
1
0
0
1
T1
- 80
-1
d
d T2 = c
480
2 J K
T3
The augmented matrix [A b] for the above set is
c
1
0
0
1
-1
2
- 80
d
480
Note that the rst two columns form a 2 2 identity matrix. This indicates that
the corresponding equations can be solved directly for T1 and T2 in terms of T3.
We can always reduce an underdetermined set to such a form by multiplying
the set’s equations by suitable factors and adding the resulting equations to eliminate an unknown variable. The MATLAB rref function provides a procedure
for reducing an equation set to this form, which is called the reduced row echelon
form. Its syntax is rref([A b]). Its output is the augmented matrix [C d] that
corresponds to the equation set Cx d. This set is in reduced row echelon form.
Three Equations in Three Unknowns, Continued
EXAMPLE 8.3–3
The following underdetermined equation set was analyzed in Example 8.3–1. There it
was shown that an in nite number of solutions exist. Use the rref function to obtain the
solutions.
2x1 - 4x2 + 5x3 = - 4
-4x1 - 2x2 + 3x3 =
4
2x1 + 6x2 - 8x3 =
0
■ Solution
The MATLAB session is
>>A = [2,-4,5;-4,-2,3;2,6,-8];
>>b = [-4;4;0];
>>rref([A, b])
ans =
1
0
-0.1
-1.2000
0
1
-1.3
0.4000
0
0
0
0
The answer corresponds to the augmented matrix [C d], where
[C d] =
1
0
J
0
0
1
0
-0.1
-1.3
0
-1.2
0.4
K
0
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347
This matrix corresponds to the matrix equation Cx d, or
x1 0x2 0.1x3 1.2
0x1 x2 1.3x3 0.4
0x1 0x2 0x3 0
These can be easily solved for x1 and x2 in terms of x3 as follows: x1 0.1x3 1.2, x2 1.3x3 0.4. This is the general solution to the problem, where x3 is taken to be the arbitary
variable.
Supplementing Underdetermined Systems
Often the linear equations describing the application are underdetermined because
not enough information has been speci ed to determine unique values of the
unknowns. In such cases we might be able to include additional information,
objectives, or constraints to nd a unique solution. We can use the rref command
to reduce the number of unknown variables in the problem, as illustrated in the
next two examples.
Production Planning
EXAMPLE 8.3–4
The following table shows how many hours reactors A and B need to produce 1 ton each
of the chemical products 1, 2, and 3. The two reactors are available for 40 and 30 hr per
week, respectively. Determine how many tons of each product can be produced each
week.
Hours
Product 1
Product 2
Product 3
Reactor A
Reactor B
5
3
3
3
3
4
■ Solution
Let x, y, and z be the number of tons each of products 1, 2, and 3 that can be produced in
one week. Using the data for reactor A, the equation for its usage in one week is
5x + 3y + 3z = 40
The data for reactor B gives
3x + 3y + 4z = 30
This system is underdetermined. The matrices for the equation Ax b are
5
A = c
3
3
3
3
d
4
40
b = c d
30
x
x = y
JzK
Here the rank(A) rank([A b]) 2, which is less than the number of unknowns. Thus
an in nite number of solutions exist, and we can determine two of the variables in terms
of the third.
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Using the rref command rref([A b]), where A = [5,3,3;3,3,4] and
b = [40;30], we obtain the following reduced echelon augmented matrix:
c
1
0
0
1
-0.5
1.8333
5
d
5
This matrix gives the reduced system
x - 0.5z = 5
y + 1.8333z = 5
which can be easily solved as follows:
x = 5 + 0.5z
(8.3–4)
y = 5 - 1.8333z
(8.3–5)
where z is arbitrary. However, z cannot be completely arbitrary if the solution is to be
meaningful. For example, negative values of the variables have no meaning here; thus we
require that x Ú 0, y Ú 0, and z Ú 0. Equation (8.3–4) shows that x Ú 0 if z Ú - 10.
From Equation (8.3–5), y Ú 0 implies that z … 5>1.8333 = 2.727. Thus valid solutions
are those given by Equations (8.3–4) and (8.3–5), where 0 … z … 2.737 tons. The choice
of z within this range must be made on some other basis, such as pro t.
For example, suppose we make a pro t of $400, $600, and $100 per ton for products
1, 2, and 3, respectively. Then our total pro t P is
P = 400x + 600y + 100z
= 400(5 + 0.5z) + 600(5 - 1.8333z) + 100z
= 5000 - 800z
Thus to maximize pro t, we should choose z to be the smallest possible value, namely,
z = 0. This choice gives x = y = 5 tons.
However, if the pro ts for each product were $3000, $600, and $100, the total pro t
would be P = 18 000 + 500z . Thus we should choose z to be its maximum, namely,
z 2.727 tons. From Equations (8.3–4) and (8.3–5), we obtain x 6.36 and y 0 tons.
Traf c Engineering
EXAMPLE 8.3–5
A traf c engineer wants to know if measurements of traf c ow entering and leaving a
road network are suf cient to predict the traf c ow on each street in the network. For
example, consider the network of one-way streets shown in Figure 8.3–2. The numbers
shown are the measured traf c ows in vehicles per hour . Assume that no vehicles park
anywhere within the network. If possible, calculate the traf c ows f1, f2, f3, and f4. If this
is not possible, suggest how to obtain the necessary information.
■ Solution
The ow into intersection 1 must equal the ow out of the intersection. This gives
100 + 200 = f1 + f4
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8.3
200
300
f1
100
200
2
1
f4
f2
4
300
Underdetermined Systems
3
f3
400
600
500
Figure 8.3–2 A network of one-way streets.
Similarly, for the other three intersections, we have
f1 + f2 = 300 + 200
600 + 400 = f2 + f3
f3 + f4 = 300 + 500
Putting these in the matrix form Ax b, we obtain
1
1
A = ≥
0
0
0
1
1
0
0
0
1
1
300
500
b = ≥
¥
1000
800
1
0
¥
0
1
f1
f2
x = ≥ ¥
f3
f4
First, check the ranks of A and [A b], using the MATLAB rank function. Both
have a rank of 3, which is 1 less than the number of unknowns, so we can determine
three of the unknowns in terms of the fourth. Thus we cannot determine all the traf c
ows based on the given measurements.
Using the rref([A b]) function produces the reduced augmented matrix
1
0
≥
0
0
0
1
0
0
0
0
1
0
1
-1
1
0
300
200
¥
800
0
which corresponds to the reduced system
f1 + f4 = 300
f2 - f4 = 200
f3 + f4 = 800
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These can be solved easily as follows: f1 300 f4, f2 200 f4, and f3 800 f4. If
we could measure the ow on one of the internal roads, say f4, then we could compute the
other ows. So we recommend that the engineer arrange to have this additional measurement made.
Test Your Understanding
T8.3–2 Use the rref, pinv, and the left division methods to solve the following set.
3x1 + 5x2 + 6x3 =
6
8x1 - x2 + 2x3 =
1
5x1 - 6x2 - 4x3 = - 5
(Answer: There are an in nite number of solutions. The result obtained
with the rref function is x1 0.2558 0.3721x3, x2 1.0465 0.9767x3, x3 arbitrary. The pinv function gives x1 0.0571, x2 =
0.5249, x3 0.5340. The left division method generates an error
message.)
T8.3–3 Use the rref, pinv, and left division methods to solve the
following set.
3x1 + 5x2 + 6x3 = 4
x1 - 2x2 - 3x3 = 10
(Answer: There are an in nite number of solutions. The result obtained
with the rref function is x1 0.2727x3 5.2727, x2 1.3636x3 2.2626, x3 arbitrary. The solution obtained with left division is x1 4.8000, x2 0, x3 1.7333. The one obtained with the pseudoinverse
method is x1 4.8394, x2 0.1972, x3 1.5887.)
8.4 Overdetermined Systems
An overdetermined system is a set of equations that has more independent equations than unknowns. Some overdetermined systems have exact solutions, and
they can be obtained with the left division method x = A\b. For other overdetermined systems, no exact solution exists; in some of these cases, the left division method does not yield an answer, while in other cases the left division
method gives an answer that satis es the equation set only in a “least-squares”
sense. We will show what this means in the next example. When MATLAB gives
an answer to an overdetermined set, it does not tell us whether the answer is the
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351
exact solution. We must determine this information ourselves, and we will now
show how to do this.
The Least-Squares Method
EXAMPLE 8.4–1
Suppose we have the following three data points, and we want to nd the straight line
y c1x c2 that best ts the data in some sense.
x
y
0
5
10
2
6
11
(a) Find the coef cients c1 and c2 using the least-squares criterion. (b) Find the coef cients by using the left division method to solve the three equations (one for each data
point) for the two unknowns c1 and c2. Compare with the answer from part (a).
■ Solution
(a) Because two points de ne a straight line, unless we are extremely lucky , our three data
points will not lie on the same straight line. A common criterion for obtaining the straight
line that best ts the data is the least-squares criterion. According to this criterion, the line
that minimizes J, the sum of the squares of the vertical differences between the line and
the data points, is the “best” t. Here J is
i=3
J = a (c1xi + c2 - yi)2 = (0c1 + c2 - 2)2 + (5c1 + c2 - 6)2 + (10c1 + c2 - 11)2
i=1
If you are familiar with calculus, you know that the values of c1 and c2 that minimize J
are found by setting the partial derivatives J/ c1 and J/ c2 equal to zero.
J
= 250c1 + 30c2 - 280 = 0
c1
J
= 30c1 + 6c2 - 38 = 0
c2
The solution is c1 0.9 and c2 11/6. The best straight line in the least-squares sense is
y 0.9x 11/6.
(b) Evaluating the equation y c1x c2 at each data point gives the following
three equations, which are overdetermined because there are more equations than
unknowns.
0c1 + c2 = 2
(8.4–1)
5c1 + c2 = 6
(8.4–2)
10c1 + c2 = 11
(8.4–3)
LEAST-SQUARES
METHOD
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These equations can be written in the matrix form Ax b as follows.
Ax =
where
0
5
J 10
1
2
c1
0 c d =
6 = b
J K
c
1K 2
11
0
[A b] =
5
J
10
1
1
1
2
6
K
11
To use left division, the MATLAB session is
>>A = [0,1;5,1;10,1];
>>b = [2;6;11];
>>rank(A)
ans =
2
>>rank([A, b])
ans =
3
>>x = A\b
x =
0.9000
1.8333
>>A*x
ans =
1.833
6.333
10.8333
This result for x agrees with the least-squares solution obtained previously: c1 0.9,
c2 11/6 1.8333. The rank of A is 2, but the rank of [A b] is 3, so no exact solution
exists for c1 and c2. Note that A*x gives the y values generated by the line y 0.9x 1.8333 at the x data values x 0, 5, 10. These are different from the right-hand sides of
the original three equations (8.4–1) through (8.4–3). This is not unexpected, because the
least-squares solution is not an exact solution of the equations.
Some overdetermined systems have an exact solution. The left division
method sometimes gives an answer for overdetermined systems, but it does not
indicate whether the answer is the exact solution. We need to check the ranks of
A and [A b] to know if the answer is the exact solution. The next example illustrates this situation.
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An Overdetermined Set
EXAMPLE 8.4–2
Solve the following equations and discuss the solution for two cases: c 9 and c 10.
x1 + x2 = 1
x1 + 2x2 = 3
x1 + 5x2 = c
■ Solution
The coef cient matrix and the augmented matrix for this problem are
A =
353
1
1
J1
1
2
5K
[A b] =
1
1
J1
1
2
5
1
3
cK
Making the computations in MATLAB, we nd that for c 9, rank (A) rank([A b]) 2.
Thus the system has a solution, and because the number of unknowns (2) equals the rank
of A, there is a unique solution. The left division method A\b gives this solution, which is
x1 1 and x2 2.
For c 10 we nd that rank(A) 2, but rank([A b]) 3. Because rank(A) rank([A b]), there is no solution. However, the left division method A\b gives x1 1.3846
and x2 2.2692, which is not an exact solution! This can be veri ed by substituting these
values into the original equation set. This answer is the solution to the equation set in a
least-squares sense. That is, these values are the values of x1 and x2 that minimize J, the sum
of the squares of the differences between the equations’ left- and right-hand sides.
J = (x1 + x2 - 1)2 + (x1 + 2x2 - 3)2 + (x1 + 5x2 - 10)2
To interpret MATLAB answers correctly for an overdetermined system, rst
check the ranks of A and [A b] to see if an exact solution exists; if one does not exist,
then we know that the left division answer is a least-squares solution. In Section 8.5
we develop a general-purpose program that checks the ranks and solves a general
set of linear equations.
Test Your Understanding
T8.4–1 Solve the following set.
x1 - 3x2 =
2
3x1 + 5x2 =
7
70x1 - 28x2 = 153
(Answer: There is a unique solution: x1 2.2143, x2 0.0714, which is
given by the left division method.)
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T8.4–2 Show why there is no solution to the following set.
x1 - 3x2 =
2
3x1 + 5x2 =
7
5x1 - 2x2 = - 4
8.5 A General Solution Program
In this chapter you saw that the set of linear algebraic equations Ax ⴝ b with m
equations and n unknowns has solutions if and only if (1) rank[A] rank[A b].
Let r rank[A]. If condition (1) is satis ed and if r n, then the solution is
unique. If condition (1) is satis ed but r
n, an in nite number of solutions
exist; in addition, r unknown variables can be expressed as linear combinations
of the other n r unknown variables, whose values are arbitrary. In this case we
can use the rref command to nd the relations between the variables. The pseudocode in Table 8.5–1 can be used to outline an equation solver program before
writing it.
A condensed owchart is shown in Figure 8.5–1. From this chart or the pseudocode, we can develop the script le shown in Table 8.5–2. The program uses
the given arrays A and b to check the rank conditions; the left division method to
obtain the solution, if one exists; and the rref method if there are an in nite
number of solutions. Note that the number of unknowns equals the number of
columns in A, which is given by size_A (2), the second element in size_A.
Note also that the rank of A cannot exceed the number of columns in A.
Test Your Understanding
T8.5–1 Type in the script le lineq.m given in Table 8.5–2 and run it for the
following cases. Hand-check the answers.
a. A = [1,-1;1,1], b = [3;5]
b. A = [1,-1;2,-2], b = [3;6]
c. A = [1,-1;2,-2], b = [3;5]
Table 8.5–1 Pseudocode for the linear equation solver
If the rank of A equals the rank of [A b], then
determine whether the rank of A equals the number of unknowns. If so, there is a
unique solution, which can be computed using left division. Display the results
and stop.
Otherwise, there are an in nite number of solutions, which can be found from
the augmented matrix. Display the results and stop.
Otherwise (if the rank of A does not equal the rank of [A b]), there are no solutions.
Display this message and stop.
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Table 8.5–2 MATLAB program to solve linear equations
% Script le lineq.m
% Solves the set Ax = b, given A and b.
% Check the ranks of A and [A b].
if rank(A) == rank([A b])
% The ranks are equal.
size_A = size(A);
% Does the rank of A equal the number of unknowns?
if rank(A) == size_A(2)
% Yes. Rank of A equals the number of unknowns.
disp(‘There is a unique solution, which is:’)
x = A\b % Solve using left division.
else
% Rank of A does not equal the number of unknowns.
disp(‘There is an in nite number of solutions.’)
disp(‘The augmented matrix of the reduced system is:’)
rref([A b]) % Compute the augmented matrix.
end
else
% The ranks of A and [A b] are not equal.
disp(‘There are no solutions.’)
end
A, b
rank (A) = rank ([A b])
?
No
Yes
No
rank (A) = # of unknowns
?
Display message:
No solutions exist.
Yes
Infinite # of solutions exist.
Compute augmented matrix
using rref command.
Unique solution exists.
Compute it with A\b.
Display answer.
Display answer.
Stop
Figure 8.5–1 Flowchart illustrating a program to solve linear equations.
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8.6 Summary
If the number of equations in the set equals the number of unknown variables,
MATLAB provides two ways of solving the equation set Ax ⴝ b: the matrix
inverse method, x = inv(A)*b, and the matrix left division method, x = A\b.
If MATLAB does not generate an error message when you use one of these
methods, then the set has a unique solution. You can always check the solution
for x by typing Ax to see if the result is the same as b. If you receive an error
message, the set is underdetermined (even though it may have an equal number
of equations and unknowns), and either it does not have a solution, or it has more
than one solution.
For underdetermined sets, MATLAB provides three ways of dealing with
the equation set Ax ⴝ b (note that the matrix inverse method will never work
with such sets):
1. The matrix left division method (which gives one speci c solution, but not
the general solution).
2. The pseudoinverse method. Solve for x by typing x = pinv(A)*b. This
gives the minimum norm solution.
3. The reduced row echelon form (RREF) method. This method uses the
MATLAB command rref to obtain a general solution for some of the unknowns in terms of the other unknowns.
The four methods are summarized in Table 8.6–1. You should be able to determine whether a unique solution, an in nite number of solutions, or no solution
exists. You can do this by applying the existence and uniqueness test given on
page 335.
Some overdetermined systems have exact solutions, and they can be obtained
with the left division method, but the method does not indicate that the solution is
exact. To determine this, rst check the ranks of A and [A b] to see if a solution
exists; if one does not exist, then we know that the left division solution is a leastsquares answer.
Table 8.6–1 Matrix functions and commands for solving linear equations
Function
Description
det(A)
inv(A)
pinv(A)
rank(A)
rref([A b])
Computes the determinant of the array A.
Computes the inverse of the matrix A.
Computes the pseudoinverse of the matrix A.
Computes the rank of the matrix A.
Computes the reduced row echelon form corresponding to the
augmented matrix [A b].
Solves the matrix equation Ax ⴝ b using the matrix inverse.
Solves the matrix equation Ax ⴝ b using left division.
x inv(A)*b
x A\b
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Key Terms with Page References
Minimum norm solution, 342
Overdetermined system, 350
Pseudoinverse method, 342
Reduced row echelon form, 345
Singular matrix, 333
Statically indeterminate, 343
Subdeterminant, 335
Underdetermined system, 341
Augmented matrix, 335
Euclidean norm, 343
Gauss elimination, 335
Homogeneous equations, 335
Ill-conditioned equations, 333
Least-squares method, 351
Left division method, 335
Matrix inverse, 332
Matrix rank, 334
Problems
You can nd the answers to problems marked with an asterisk at the end of the text.
Section 8.1
1.
Solve the following problems using matrix inversion. Check your solutions
by computing A1A.
a. 2x + y = 5
3x - 9y = 7
b. - 8x - 5y = 4
- 2x + 7y = 10
2.*
c.
12x - 5y = 11
- 3x + 4y + 7x3 = - 3
6x + 2y + 3x3 = 22
d.
6x - 3y + 4x3 = 41
12x + 5y - 7x3 = - 26
- 5x + 2y + 6x3 = 16
a. Solve the following matrix equation for the matrix C.
A(BC + A) ⴝ B
b. Evaluate the solution obtained in part a for the case
A = c
7
-2
9
d
4
B = c
4
7
-3
d
6
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3.
Use MATLAB to solve the following problems.
a. -2x + y = - 5
- 2x + y = 3
b. - 2x + y = 3
- 8x + 4y = 12
c. -2x + y = - 5
- 2x + y = - 5.00001
d.
x1 + 5x2 - x3 + 6x4 = 19
2x1 - x2 + x3 - 2x4 =
7
- x1 + 4x2 - x3 + 3x4 = 30
3x1 - 7x2 - 2x3 + x4 = - 75
Section 8.2
4.
The circuit shown in Figure P4 has ve resistances and one applied voltage.
Kirchhoff’s voltage law applied to each loop in the circuit shown gives
␷ - R2i2 - R4i4 = 0
- R2i2 + R1i1 + R3i3 = 0
- R4i4 - R3i3 + R5i5 = 0
i1
R1
R2
R3
i2
+
v
i4
–
i3
R4
R5
i5
i6
Figure P4
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Problems
Conservation of charge applied at each node in the circuit gives
i6 = i1 + i2
i2 + i3 = i4
i1 = i3 + i5
i4 + i5 = i6
5.*
a. Write a MATLAB script le that uses given values of the applied voltage
␷ and the values of the ve resistances and solves for the six currents.
b. Use the program developed in part a to nd the currents for the case
where R1 1, R2 5, R3 2, R4 10, R5 5 k , and ␷ 100 V.
(1 k 1000 .)
a. Use MATLAB to solve the following equations for x, y, and z as
functions of the parameter c.
x - 5y - 2z = 11c
6x + 3y + z = 13c
7x + 3y - 5z = 10c
b. Plot the solutions for x, y, and z versus c on the same plot, for
10 c 10.
6.
Fluid ows in pipe networks can be analyzed in a manner similar to that
used for electric resistance networks. Figure P6 shows a network with
three pipes. The volume ow rates in the pipes are q1, q2, and q3. The
pressures at the pipe ends are pa, pb, and pc. The pressure at the junction is
p1. Under certain conditions, the pressure– ow rate relation in a pipe has
the same form as the voltage-current relation in a resistor. Thus, for the
three pipes, we have
q1 =
1
(p - p1)
R1 a
q2 =
1
( p - pb)
R2 1
q3 =
1
(p - pc)
R3 1
where the Ri are the pipe resistances. From conservation of mass,
q1 q2 q3.
a. Set up these equations in a matrix form Ax b suitable for solving for
the three ow rates q1, q2, and q3 and the pressure p1, given the values
of pressures pa, pb, and pc and the values of resistances R1, R2, and R3.
Find the expressions for A and b.
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CHAPTER 8 Linear Algebraic Equations
pb
R2
q2
R1
pa
q1
p1
R3
q3
pc
(a)
pb
R2
R1
pa
p1
R3
pc
(b)
Figure P6
b. Use MATLAB to solve the matrix equations obtained in part a for the
case where pa 4320 lb/ft2, pb 3600 lb/ft2, and pc 2880 lb/ft2.
These correspond to 30, 25, and 20 psi, respectively (1 psi 1 lb/in2,
and atmospheric pressure is 14.7 psi). Use the resistance values R1 10 000, R2 14 000 lb sec/ft5. These values correspond to fuel oil
owing through pipes 2 ft long, with 2- and 1.4-in. diameters, respectively. The units of the answers are ft3/sec for the ow rates and lb/ft 2
for pressure.
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Problems
y
Han d
θ2
L2
L1
Elbo w Mot or
θ1
B ase Mo tor
x
(a)
Path of
Hand
Fi nish
Start
(b)
Figure P7
7.
Figure P7 illustrates a robot arm that has two “links” connected by two
“joints”—a shoulder or base joint and an elbow joint. There is a motor at
each joint. The joint angles are 1 and 2. The (x, y) coordinates of the
hand at the end of the arm are given by
x = L1 cos 1 + L2 cos (1 + 2 )
y = L1 sin 1 + L2 sin (1 + 2)
where L1 and L2 are the lengths of the links.
Polynomials are used for controlling the motion of robots. If we start the
arm from rest with zero velocity and acceleration, the following
polynomials are used to generate commands to be sent to the joint
motor controllers
1(t) = 1(0) + a1t3 + a2t4 + a3t5
2(t) = 2(0) + b1t3 + b2t4 + b3t5
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CHAPTER 8 Linear Algebraic Equations
where 1(0) and 2(0) are the starting values at time t 0. The angles
1(tf) and 2(tf) are the joint angles corresponding to the desired destination of the arm at time tf . The values of 1(0), 2(0), 1(tf), and 2(tf) can
be found from trigonometry, if the starting and ending (x, y) coordinates
of the hand are speci ed.
a. Set up a matrix equation to be solved for the coef cients a1, a2, and a3,
given values for 1(0), 1(tf), and tf . Obtain a similar equation for the
coef cients b1, b2, and b3.
b. Use MATLAB to solve for the polynomial coef cients given the values
tf 2 sec, 1(0) 19, 2(0) 44, 1(tf) 43, and 2(tf) 151.
(These values correspond to a starting hand location of x 6.5, y 0 ft
and a destination location of x 0, y 2 ft for L1 4 and L2 3 ft.)
c. Use the results of part b to plot the path of the hand.
8.*
Engineers must be able to predict the rate of heat loss through a building
wall to determine the heating system requirements. They do this by using
the concept of thermal resistance R, which relates the heat ow rate q
through a material to the temperature difference T across the material:
q T/R. This relation is like the voltage-current relation for an electric
resistor: i v/R. So the heat ow rate plays the role of electric current, and
the temperature difference plays the role of the voltage difference. The
SI unit for q is the watt (W), which is 1 joule/second (J/s).
The wall shown in Figure P8 consists of four layers: an inner layer of
plaster/lathe 10 mm thick, a layer of ber glass insulation 125 mm thick, a
Lathe
Insulation
Wood
Brick
Inside
Air
Outside
Air
Ti
T1
T2
T3
To
T3
To
(a)
Ti
T1
R1
T2
R3
R2
(b)
Figure P8
R4
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Problems
layer of wood 60 mm thick, and an outer layer of brick 50 mm thick. If
we assume that the inner and outer temperatures Ti and To have remained
constant for some time, then the heat energy stored in the layers is
constant, and thus the heat ow rate through each layer is the same.
Applying conservation of energy gives the following equations.
q =
1
1
1
1
(Ti - T1) =
(T1 - T2) =
(T2 - T3) =
(T - To)
R1
R2
R3
R4 3
The thermal resistance of a solid material is given by R D/k, where D is
the material thickness and k is the material’s thermal conductivity. For the
given materials, the resistances for a wall area of 1 m2 are R1 0.036,
R2 4.01, R3 0.408, and R4 0.038 K/W.
Suppose that Ti 20C and To 10C. Find the other three temperatures
and the heat loss rate q, in watts. Compute the heat loss rate if the wall’s area
is 10 m2.
9.
The concept of thermal resistance described in Problem 8 can be used to nd
the temperature distribution in the at square plate shown in Figure P9(a).
Ta
T1
T2
T3
T4
Tb
( a)
T2
T1
Ta
R
R
R
R
R
T3
T4
(b )
Figure P9
R
Tb
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CHAPTER 8 Linear Algebraic Equations
The plate’s edges are insulated so that no heat can escape, except at two
points where the edge temperature is heated to Ta and Tb, respectively.
The temperature varies through the plate, so no single point can describe
the plate’s temperature. One way to estimate the temperature distribution
is to imagine that the plate consists of four subsquares and to compute
the temperature in each subsquare. Let R be the thermal resistance of the
material between the centers of adjacent subsquares. Then we can think
of the problem as a network of electric resistors, as shown in part (b) of
the gure. Let qij be the heat ow rate between the points whose temperatures are Ti and Tj. If Ta and Tb remain constant for some time, then the
heat energy stored in each subsquare is constant also, and the heat ow
rate between each subsquare is constant. Under these conditions, conservation of energy says that the heat ow into a subsquare equals the heat
ow out. Applying this principle to each subsquare gives the following
equations.
qa1 = q12 + q13
q12 = q24
q13 = q34
q34 + q24 = q4b
Substituting q (Ti Tj)/R, we nd that R can be canceled out of every
equation, and they can be rearranged as follows:
T1 =
1
(T + T2 + T3)
3 a
T2 =
1
(T + T4)
2 1
T3 =
1
(T + T4)
2 1
T4 =
1
(T2 + T3 + T5)
3
These equations tell us that the temperature of each subsquare is the average of the temperatures in the adjacent subsquares!
Solve these equations for the case where Ta 150C and Tb 20C.
10.
Use the averaging principle developed in Problem 9 to nd the temperature distribution of the plate shown in Figure P10, using the 3 3 grid
and the given values Ta 150C and Tb 20C.
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Problems
Ta
T1
T2
T3
T4
T5
T6
T7
T8
T9
Tb
Figure P10
Section 8.3
11.* Solve the following equations:
7x + 9y - 9z = 22
3x + 2y - 4z = 12
x 5y z 2
12.
The following table shows how many hours in process reactors A and B
are required to produce 1 ton each of chemical products 1, 2, and 3. The
two reactors are available for 35 and 40 hrs per week, respectively.
Hours
Reactor A
Reactor B
Product 1
Product 2
Product 3
6
3
2
5
10
2
Let x, y, and z be the number of tons each of products 1, 2, and 3 that can
be produced in one week.
a. Use the data in the table to write two equations in terms of x, y, and z.
Determine whether a unique solution exists. If not, use MATLAB to
nd the relations between x, y, and z.
b. Note that negative values x, y, and z have no meaning here. Find the
allowable ranges for x, y, and z.
c. Suppose the pro ts for each product are $200, $300, and $100 for
products 1, 2, and 3, respectively. Find the values of x, y, and z to
maximize the pro t.
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1
100
f1
f3
300
100
300
200
2
f2
3
f5
f4
4
5
f6
100
500
6
f7
200
200
400
Figure P13
d. Suppose the pro ts for each product are $200, $500, and $100 for
products 1, 2, and 3, respectively. Find the values of x, y, and z to
maximize the pro t.
13.
See Figure P13. Assume that no vehicles stop within the network. A
traf c engineer wants to know if the traf c ows f1, f2, . . . , f7 (in vehicles
per hour) can be computed given the measured ows shown in the gure.
If not, then determine how many more traf c sensors need to be installed,
and obtain the expressions for the other traf c ows in terms of the
measured quantities.
Section 8.4
14.* Use MATLAB to solve the following problem:
x - 3y = 2
x + 5y = 18
4x - 6y = 20
15.* Use MATLAB to solve the following problem:
x - 3y = 2
x + 5y = 18
4x - 6y = 10
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Problems
16.
a. Use MATLAB to nd the coef cients of the quadratic polynomial
y ax2 bx c that passes through the three points (x, y) (1, 4),
(4, 73), (5, 120).
b. Use MATLAB to nd the coef cients of the cubic polynomial y ax3 bx2 cx d that passes through the three points given in part a.
17.
Use the MATLAB program given in Table 8.5–2 to solve the following
problems:
a. Problem 3d
b. Problem 11
c. Problem 14
d. Problem 15
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© Donovan Reese/Getty Images/RF
Engineering in the
21st Century. . .
Rebuilding the Infrastructure
D
uring the Great Depression, many public works projects that improved
the nation’s infrastructure were undertaken to stimulate the economy and
provide employment. These projects included highways, bridges, water
supply systems, sewer systems, and electrical power distribution networks. Following World War II another burst of such activity culminated in the construction
of the interstate highway system. As we enter the 21st century, much of the infrastructure is 30 to 70 years old and is literally crumbling or not up to date. One
survey showed that more than 25 percent of the nation’s bridges are substandard.
These need to be repaired or replaced.
Rebuilding the infrastructure requires engineering methods different from
those in the past because labor and material costs are now higher and environmental and social issues have greater importance than before. Infrastructure engineers
must take advantage of new materials, inspection technology, construction techniques, and labor-saving machines.
Also, some infrastructure components, such as communications networks,
need to be replaced because they are outdated and do not have suf cient capacity
or ability to take advantage of new technology. An example is the information
infrastructure, which includes physical facilities to transmit, store, process, and
display voice, data, and images. Better communications and computer networking
technology will be needed for such improvements. Many of the MATLAB toolboxes provide advanced support for such work, including the Financial, Communications, Image Processing, Signal Processing, PDE, and Wavelet toolboxes. ■
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C H A P T E R
9
Numerical Methods
for Calculus and
Differential Equations
OUTLINE
9.1 Numerical Integration
9.2 Numerical Differentiation
9.3 First-Order Differential Equations
9.4 Higher-Order Differential Equations
9.5 Special Methods for Linear Equations
9.6 Summary
Problems
This chapter covers numerical methods for computing integrals and derivatives
and for solving ordinary differential equations. Some integrals cannot be evaluated analytically, and we need to compute them numerically with an approximate
method (Section 9.1). In addition, it is often necessary to use data to estimate rates
of change, and this requires a numerical estimate of the derivative (Section 9.2).
Finally, many differential equations cannot be solved analytically, and so we
must be able to solve them by using appropriate numerical techniques. Section 9.3
covers rst-order dif ferential equations, and Section 9.4 extends the methods to
higher-order equations. More powerful methods are available for linear equations.
Section 9.5 treats these methods.
When you have nished this chapter , you should be able to
■
■
■
Use MATLAB to numerically evaluate integrals.
Use numerical methods with MATLAB to estimate derivatives.
Use the MATLAB numerical differential equation solvers to obtain solutions.
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9.1 Numerical Integration
The integral of a function f (x) for a x b can be interpreted as the area between the f (x) curve and the x axis, bounded by the limits x a and x b. If we
denote this area by A, then we can write A as
b
A =
DEFINITE
INTEGRAL
INDEFINITE
INTEGRAL
IMPROPER
INTEGRAL
SINGULARITIES
La
f (x) dx
(9.1–1)
An integral is called a de nite integral if it has speci ed limits of integration.
Inde nite integrals have no speci ed limits. Improper integrals can have in nite
values, depending on their integration limits. For example, the following integral
can be found in most integral tables:
1
dx = ln ⱍx - 1ⱍ
Lx - 1
However, it is an improper integral if the integration limits include the point x 1.
So, even though an integral can be found in an integral table, you should examine the integrand to check for singularities, which are points at which the integrand is unde ned. The same warning applies when you are using numerical
methods to evaluate integrals.
Trapezoidal Integration
The simplest way to nd the area under a curve is to split the area into rectangles
(Figure 9.1–1a). If the widths of the rectangles are small enough, the sum of their
areas gives the approximate value of the integral. A more sophisticated method is
to use trapezoidal elements (Figure 9.1–1b). Each trapezoid is called a panel. It
is not necessary to use panels of the same width; to increase the method’s
y
y
Rectangular
Trapezoidal
y = f (x)
a
y = f (x)
b
(a)
x
a
b
x
(b)
Figure 9.1–1 Illustration of (a) rectangular and (b) trapezoidal numerical integration.
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9.1
Numerical Integration
371
accuracy, you can use narrow panels where the function is changing rapidly.
When the widths are adjusted according to the function’s behavior, the method is
said to be adaptive. MATLAB implements trapezoidal integration with the
trapz function. Its syntax is trapz(x, y), where the array y contains the
function values at the points contained in the array x. If you want the integral of
a single function, then y is a vector. To integrate more than one function, place
their values in a matrix y; trapz(x, y) will compute the integral of each
column of y.
You cannot directly specify a function to integrate with the trapz function;
you must rst compute and store the function’ s values ahead of time in an array.
Later we will discuss two other integration functions, the quad and quadl
functions, that can accept functions directly. However, they cannot handle arrays
of values. So the functions complement each other. The trapz function is
summarized in Table 9.1–1.
As a simple example of the use of the trapz function, let us compute the
integral
A =
L0
sin x dx
(9.1–2)
whose exact answer is A 2. To investigate the effect of panel width, let us rst
use 10 panels with equal widths of 兾10. The script le is
x = linspace(0,pi,10);
y = sin(x);
A = trapz(x,y)
Table 9.1–1 Basic syntax of numerical integration functions
Command
Description
dblquad(fun,a,b,c,d)
Computes the double integral of the function f (x, y) between the
limits a x b and c y d. The input fun speci es the function that computes the integrand. It must accept a vector argument x
and scalar y, and it must return a vector result.
Computes the integral of the polynomial p using an optional
user-speci ed constant of integration C.
Uses an adaptive Simpson rule to compute the integral of the function
fun between the limits a and b. The input fun, which represents the
integrand f (x), is a function handle for the integrand function. It must
accept a vector argument x and return the vector result y.
Uses Lobatto integration. The syntax is identical to quad.
Uses trapezoidal integration to compute the integral of y with
respect to x, where the array y contains the function values at the
points contained in the array x.
Computes the triple integral of the function f (x, y, z) between the
limits a x b, c y d, and e y f. The input fun
speci es the function that computes the integrand. It must accept a
vector argument x, scalar y, and scalar z, and it must return a vector
result.
polyint(p,C)
quad(fun,a,b)
quadl(fun,a,b)
trapz(x,y)
triplequad(fun,a,b,c,d,e,f)
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The answer is A 1.9797, which gives a relative error of
100(2 1.9797)兾2 1 percent. Now try 100 panels of equal width; replace the
array x with x = linspace(0,pi,100). The answer is A 1.9998 for a
relative error of 100(2 1.9998)兾2 0.01 percent. If we examine the plot of the
integrand sin x, we see that the function is changing faster near x 0 and x than near x 兾2. Thus we could achieve the same accuracy by using fewer
panels if narrower panels are used near x 0 and x .
We normally use the trapz function when the integrand is given as a table
of values. Otherwise, if the integrand is given as a function, use the quad or
quadl functions, to be introduced shortly.
Velocity from an Accelerometer
EXAMPLE 9.1–1
An accelerometer is used in aircraft, rockets, and other vehicles to estimate the vehicle’s
velocity and displacement. The accelerometer integrates the acceleration signal to produce an estimate of the velocity, and it integrates the velocity estimate to produce an
estimate of displacement. Suppose the vehicle starts from rest at time t 0, and its measured acceleration is given in the following table.
Time (s)
Acceleration (m/s2)
0
0
1
2
2
4
3
7
4
11
5
17
6
24
7
32
8
41
9
48
10
51
(a) Estimate the velocity after 10 s.
(b) Estimate the velocity at times t 1, 2, . . . , 10 s.
■ Solution
(a) The initial velocity is zero, so (0) 0. The relation between the velocity and acceleration a(t) is
10
(10) =
L0
10
a(t) dt + (0) =
L0
a(t) dt
The script le is shown below .
t = 0:10;
a = [0,2,4,7,11,17,24,32,41,48,51];
v10 = trapz(t,a)
The answer for the velocity after 10 s is v10, and it is 211.5 m/s.
(b) The following script le uses the fact that the velocity can be expressed as
tk + 1
(tk + 1) =
Ltk
a(t) dt + (tk)
where (t1) 0.
t = 0:10;
a = [0,2,4,7,11,17,24,32,41,48,51];
k = 1, 2, . . . , 10
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v(1) = 0;
for k = 1:10
v(k+1) = trapz(t(k:k+1), a(k:k+1))+v(k);
end
disp([t’,v’])
The answers are given in the following table.
Time (s)
Velocity (m/s)
0
0
1
1
2
4
3
9.5
4
18.5
5
32.5
6
53
7
81
8
117
9
162
10
211.5
Test Your Understanding
T9.1–1 Modify the script le given in part (b) of Example 9.1–1 to estimate the
displacement at times t 1, 2, . . . , 10 s. (Partial answer: The displacement after 10 s is 584.25 m.)
Quadrature Functions
Another approach to numerical integration is Simpson’s rule, which divides the integration range b a into an even number of sections and uses a different quadratic
function to represent the integrand for each panel. A quadratic function has three
parameters, and Simpson’s rule computes these parameters by requiring that the
quadratic pass through the function’s three points corresponding to the two adjacent
panels. To obtain greater accuracy, we can use polynomials of degree higher than 2.
The MATLAB function quad implements an adaptive version of Simpson’s
rule. The quadl function is based on an adaptive Lobatto integration method,
where the letter “l” in quadl stands for Lobatto. The term quad is an abbreviation of quadrature, which is an old term for the process of measuring areas. Some
writers distinguish between the terms quadrature and integration and reserve
integration to mean numerical integration of ordinary differential equations. We
will not make that distinction.
The function quad(fun,a,b) computes the integral of the function fun
between the limits a and b. The input fun, which represents the integrand f (x),
is either a function handle of the integrand function (the preferred method) or the
name of the function as a character string (i.e., placed in single quotes). The function y f (x) must accept a vector argument x and must return the vector result y.
The basic syntax of quadl is identical and is summarized in Table 9.1–1.
To illustrate, let us compute the integral given in Equation (9.1–2). The
session consists of one command: A = quad(@sin, 0, pi) or A =
quad(‘sin’, 0, pi). The answer given by MATLAB is A 2.0000, which is
correct to four decimal places. We use quad1 the same way.
Because the quad and quadl functions call the integrand function using
vector arguments, you must always use array operations when de ning the function. The following example shows how this is done.
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Evaluation for Fresnel’s Cosine Integral
EXAMPLE 9.1–2
Some simple-looking integrals cannot be evaluated in closed form. An example is
Fresnel’s cosine integral
b
A =
cos x2 dx
L0
(9.1–3)
(a) Demonstrate two ways to compute the integral when the upper limit is b = 12.
(b) Demonstrate the use of a nested function to compute the more general integral
b
A =
L0
cos xn dx
(9.1–4)
for n 2 and for n 3.
Solution
(a) The integrand cos x2 obviously does not contain any singularities that might cause problems for the integration function. We demonstrate two ways to use the quad function.
1. With a function le: De ne the integrand with a user -de ned function as shown by
the following function le.
function c2 = cossq(x)
c2 = cos(x.^2);
2.
The quad function is called as follows: A = quad(@cossq,0,sqrt(2*pi)).
The result is A 0.6119.
With an anonymous function (anonymous functions are discussed in Section 3.3):
The session is
>>cossq = @(x)cos(x.^2);
>>A = quad(cossq,0,sqrt(2*pi))
A =
0.6119
The two lines can be combined into one as follows:
A = quad(@(x)cos(x.^2),0,sqrt(2*pi))
The advantage of using an anonymous function is that you need not create and save
a function le. However , for complicated integrand functions, using a function le is
preferable.
(b) Because quad requires that the integrand function have only one argument, the
following code will not work.
>>cossq = @(x)cos(x.^n);
>>n = 2;
>>A = quad(cossq,0,sqrt(2*pi))
??? Unde ned function or variable ‘n’.
Instead we will use parameter passing with a nested function (nested functions are discussed in Section 3.3). First create and save the following function.
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function A = integral_n(n)
A = quad(@cossq_n,0,sqrt(2*pi));
% Nested function
function integrand = cossq_n(x)
integrand = cos(x.^n);
end
end
The session for n 2 and n 3 is as follows.
>>A = integral_n(2)
A =
0.6119
>>A = integral_n(3)
A =
0.7734
The quad functions have some optional arguments for analyzing and adjusting the algorithm’s ef ciency and accuracy . Type help quad for details.
Test Your Understanding
T9.1–2 Use both the quad and quadl functions to compute the integral
5
A =
1
dx
L2 x
and compare the answers with that obtained from the closed-form
solution, which is A 0.9163.
Polynomial Integration
MATLAB provides the polyint function to compute the integral of a polynomial. The syntax q = polyint(p, C) returns a polynomial q representing the integral of polynomial p with a user-speci ed scalar constant of
integration C. The elements of the vector p are the coef cients of the polynomial,
arranged in descending powers. The syntax polyint(p) assumes the constant
of integration C is zero.
For example, the integral of 12x3 9x2 8x 5 is obtained from q =
polyint([12,9,8,5], 10). The answer is q = [3, 3, 4, 5, 10],
which corresponds to 3x4 3x3 4x2 5x 10. Because polynomial integrals
can be obtained from a symbolic formula, the polyint function is not a numerical
integration operation.
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Double Integrals
The function dblquad computes double integrals. Consider the integral
d
A =
b
Lc La
f (x, y) dx dy
The basic syntax is
A = dblquad(fun, a, b, c, d)
where fun is the handle to a user-de ned function that de nes the integrand
f (x, y). The function must accept a vector x and a scalar y, and it must return a
vector result, so the appropriate array operations must be used. The extended
syntax enables the user to adjust the accuracy and to use quadl or a user-de ned
quadrature routine. See the MATLAB Help for details.
For example, using an anonymous function to compute the integral
1
A =
L0 L1
3
xy2 dx dy
you type
>>fun = @(x,y)x.*y^2;
>>A = dblquad(fun, 1, 3, 0, 1)
The answer is A 1.3333.
The preceding integral is carried out over the rectangular region speci ed by
1 x 3, 0 y 1. Some double integrals are speci ed over a nonrectangular region. These problems can be handled by a transformation of variables. You
can also use a rectangular region that encloses the nonrectangular region and
force the integrand to be zero outside of the nonrectangular region, by using the
MATLAB relational operators, for example. See Problem 15. The following
example illustrates the former approach.
Double Integral over a Nonrectangular Region
EXAMPLE 9.1–3
Compute the integral
A =
L LR
(x - y)4(2x + y)2 dx dy
over the region R bounded by the lines
x - y = 1
2x + y = 2
Solution
We must convert the integral into one that is speci ed over a rectangular region. To do
this, let u x – y and 2x y. Thus, using the Jacobian, we obtain
dx dy = `
x>u
y>u
1
x>
1>3 1>3
` du d = `
` du d = du d
y>
-2>3 1>3
3
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Then the region R is speci ed as a rectangular region in terms of u and . Its boundaries
are given by u 1 and 2, and the integral becomes
2
A =
1
1
u42 du d
3 L-2 L-1
and the MATLAB session is
>>fun = @(u,v)u.^4*v^2;
>>A = (1/3)*dblquad(fun, -1, 1, -2, 2)
The answer is A 0.7111.
Triple Integrals
The function triplequad computes triple integrals. Consider the integral
f
A =
d
b
f (x, y, z) dx dy dz
Le Lc La
The basic syntax is
A = triplequad(fun, a, b, c, d, e, f)
where fun is the handle to a user-de ned function that de nes the integrand
f (x, y, z). The function must accept a vector x, a scalar y, and a scalar z, and it
must return a vector result, so the appropriate array operations must be used. The
extended syntax enables the user to adjust the accuracy and to use quadl or a
user-de ned quadrature routine. See the MA TLAB Help for details. For example, to compute the integral
2
A =
2
L1 L0 L1
3
a
xy - y2
b dx dy dz
z
You type
>>fun = @(x,y,z)(x*y-y^2)/z;
>>A = triplequad(fun, 1, 3, 0, 2, 1, 2)
The answer is A 1.8484.
9.2 Numerical Differentiation
The derivative of a function can be interpreted graphically as the slope of the
function. This interpretation leads to various methods for computing the derivative of a set of data. Figure 9.2–1 shows three data points that represent a
function y(x). Recall that the de nition of the derivative is
dy
= lim
x:0
dx
y
x
(9.2–1)
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True Slope
y3
y = f (x)
y2
B
y1
C
A
Δx
x1
Δx
x2
x3
Figure 9.2–1 Illustration of methods for estimating the
derivative dy/dx.
The success of numerical differentiation depends heavily on two factors: the
spacing of the data points and the scatter present in the data due to measurement
error. The greater the spacing, the more dif cult it is to estimate the derivative.
We assume here that the spacing between the measurements is regular; that is,
x3 x2 x2 x1 x. Suppose we want to estimate the derivative dy兾dx at the
point x2. The correct answer is the slope of the straight line passing through the
point (x2, y2); but we do not have a second point on that line, so we cannot nd
its slope. Therefore, we must estimate the slope by using nearby data points. One
estimate can be obtained from the straight line labeled A in the gure. Its slope is
mA =
BACKWARD
DIFFERENCE
(9.2–2)
This estimate of the derivative is called the backward difference estimate, and it
is actually a better estimate of the derivative at x x1 ( x)兾2 than at x x2.
Another estimate can be obtained from the straight line labeled B. Its slope is
mB =
FORWARD
DIFFERENCE
y2 - y1
y2 - y1
=
x2 - x1
x
y3 - y2
y3 - y2
=
x3 - x2
x
(9.2–3)
This estimate is called the forward difference estimate, and it is a better estimate
of the derivative at x x2 ( x)兾2 than at x x2. Examining the plot, you
might think that the average of these two slopes would provide a better estimate
of the derivative at x x2, because the average tends to cancel out the effects of
measurement error. The average of mA and mB is
mC =
mA + mB
y3 - y2
y3 - y1
1 y2 - y1
= a
+
b =
2
2
x
x
2 x
(9.2–4)
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This is the slope of the line labeled C, which connects the rst and third data
points. This estimate of the derivative is called the central difference estimate.
The diff Function
MATLAB provides the diff function to use for computing derivative estimates. Its syntax is d = diff(x), where x is a vector of values, and the result
is a vector d containing the differences between adjacent elements in x. That is,
if x has n elements, d will have n – 1 elements, where d [x(2) – x(1), x(3) –
x(2), . . . , x(n) – x(n – 1)]. For example, if x = [5, 7, 12, –20], then
diff(x) returns the vector [2, 5, –32]. The derivative dy兾dx can be estimated from diff(y)./diff(x).
The following script le implements the backward dif ference and central
difference methods for arti cial data generated from a sinusoidal signal that is
measured 51 times during one half-period. The measurement error is uniformly
distributed between –0.025 and 0.025.
x = 0:pi/50:pi;
n = length(x);
% Data-generation function with +/–0.025 random error.
y = sin(x)+.05*(rand(1,51)–0.5);
% Backward difference estimate of dy/dx.
d1 = diff(y)./diff(x);
subplot(2,1,1)
plot(x(2:n),d1,x(2:n),d1,’o’)
% Central difference estimate of dy/dx.
d2 = (y(3:n)–y(1:n–2))./(x(3:n)–x(1:n–2));
subplot(2,1,2)
plot(x(2:n–1),d2,x(2:n–1),d2,’o’)
Test Your Understanding
T9.2–1 Modify the previous program to use the forward difference method to
estimate the derivative. Plot the results, and compare with the results
from the backward and central difference methods.
Polynomial Derivatives
MATLAB provides the polyder function to compute the derivative of a polynomial. Its syntax has several forms. The basic form is d = polyder(p),
where p is a vector whose elements are the coef cients of the polynomial, arranged in descending powers. The output d is a vector containing the coef cients
of the derivative polynomial.
379
CENTRAL
DIFFERENCE
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The second syntax form is d = polyder(p1,p2). This form computes
the derivative of the product of the two polynomials p1 and p2. The third form
is [num, den] = polyder(p2,p1). This form computes the derivative of
the quotient p2/p1. The vector of coef cients of the numerator of the derivative
is given by num. The denominator is given by den.
Here are some examples of the use of polyder. Let p1 5x 2 and p2 10x2 4x – 3. Then
dp2
= 20x + 4
dx
p1 p2 = 50x3 + 40x2 - 7x - 6
d( p1 p2)
= 150x2 + 80x - 7
dx
d( p2>p1)
50x2 + 40x + 23
=
dx
25x2 + 20x + 4
These results can be obtained with the following program.
p1 = [5, 2];p2 = [10, 4, -3];
% Derivative of p2.
der2 = polyder(p2)
% Derivative of p1*p2.
prod = polyder(p1,p2)
% Derivative of p2/p1.
[num, den] = polyder(p2,p1)
The results are der2 = [20, 4], prod = [150, 80, -7], num =
[50, 40, 23], and den = [25, 20, 4].
Because polynomial derivatives can be obtained from a symbolic formula,
the polyder function is not a numerical differentiation operation.
Gradients
The gradient f of a function f (x, y) is a vector pointing in the direction of increasing values of f (x, y). It is de ned by
f =
f
f
i +
j
x
y
where i and j are the unit vectors in the x and y directions, respectively. The concept can be extended to functions of three or more variables.
In MATLAB the gradient of a set of data representing a two-dimensional
function f (x, y) can be computed with the gradient function. Its syntax is
[df_dx, df_dy] = gradient (f, dx, dy), where df_dx and df_dy
represent f兾 x and f兾 y and dx and dy are the spacing in the x and y values
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381
2
1
y
0
−1
−2
−2
−1.5
−1
−0.5
0
x
0.5
1
1.5
2
f
0.5
0
−0.5
2
1
0
−1
−2
−2
0
−1
y
x
2
2 2
Figure 9.2–2 Gradient, contour, and surface plots of the function f (x, y) = xe-(x + y ) + y2.
associated with the numerical values of f. The syntax can be extended to include
functions of three or more variables.
The following program plots the contour plot and the gradient (shown by
arrows) for the function
2
2 2
f (x, y) = xe-(x + y ) + y2
The plots are shown in Figure 9.2–2. The arrows point in the direction of
increasing f.
[x,y] = meshgrid(–2:0.25:2);
f = x.*exp(–((x–y.^2).^2+y.^2));
dx = x(1,2) – x(1,1); dy = y(2,1) – y(1,1);
[df_dx, df_dy] = gradient(f, dx, dy);
subplot(2,1,1)
contour(x,y,f), xlabel(‘x’), ylabel(‘y’), . . .
1
2
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Table 9.2–1 Numerical differentiation functions
Command
Description
d = diff(x)
Returns a vector d containing the differences between adjacent elements in the
vector x.
Computes the gradient of the function
f (x, y), where df_dx and df_dy
represent f>x and f>y, and dx and dy
are the spacing in the x and y values associated with the numerical values of f.
Returns a vector d containing the
coef cients of the derivative of the
polynomial represented by the vector p.
Returns a vector d containing the
coef cients of the polynomial that is the
derivative of the product of the
polynomials represented by p1 and p2.
Returns the vectors num and den
containing the coef cients of the
numerator and denominator polynomials
of the derivative of the quotient p2/p1,
where p1 and p2 are polynomials.
[df_dx,df_dy] =
gradient (f,dx,dy)
d = polyder(p)
d = polyder(p1,p2)
[num, den] = polyder(p2,p1)
hold on, quiver(x,y,df_dx, df_dy), hold off
subplot(2,1,2)
mesh(x,y,f),xlabel(‘x’),ylabel(‘y’),zlabel(‘f’)
LAPLACIAN
The curvature is given by the second-order derivative expression called the
Laplacian.
2
f (x, y) =
2f
2f
+
x2
y2
It can be computed with the de12 function. See the MATLAB Help for details.
The MATLAB differentiation functions discussed here are summarized in
Table 9.2–1.
9.3 First-Order Differential Equations
ORDINARY
DIFFERENTIAL
EQUATION (ODE)
INITIAL-VALUE
PROBLEM
In this section, we introduce numerical methods for solving rst-order dif ferential equations. In Section 9.4 we show how to extend the techniques to higherorder equations.
An ordinary differential equation (ODE) is an equation containing ordinary
derivatives of the dependent variable. An equation containing partial derivatives
with respect to two or more independent variables is a partial differential equation (PDE). Solution methods for PDEs are an advanced topic, and we will not
treat them in this text. In this chapter we limit ourselves to initial-value problems
(IVPs). These are problems where the ODE must be solved for a given set of
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9.3
383
values speci ed at some initial time, which is usually taken to be t = 0. Other
types of ODE problems are discussed at the end of Section 9.6.
It will be convenient to use the following abbreviated “dot” notation for
derivatives.
dy
#
y(t) =
dt
2
dy
$
y(t) = 2
dt
The free response of a differential equation, sometimes called the homogeneous
solution or the initial response, is the solution for the case where there is no forcing
function. The free response depends on the initial conditions. The forced response is
the solution due to the forcing function when the initial conditions are zero. For linear differential equations, the complete or total response is the sum of the free and
the forced responses. Nonlinear ODEs can be recognized by the fact that the dependent variable or its derivatives appear raised to
# a power or in a transcendental func#
tion. For example, the equations y = y2 and y = cos y are nonlinear.
The essence of a numerical method is to convert the differential equation
into a difference equation that can be programmed. Numerical algorithms differ
partly as a result of the speci c procedure used to obtain the dif ference equations.
It is important to understand the concept of “step size” and its effects on solution
accuracy. To provide a simple introduction to these issues, we consider the simplest numerical methods, the Euler method and the predictor-corrector method.
FREE RESPONSE
FORCED
RESPONSE
The Euler Method
The Euler method is the simplest algorithm for numerical solution of a differential equation. Consider the equations
dy
= f(t, y)
dt
y(0) = y0
(9.3–1)
where f (t, y) is a known function and y0 is the initial condition, which is the given
value of y(t) at t 0. From the de nition of the derivative,
y(t +
dy
= lim
t:0
dt
t) - y(t)
t
If the time increment t is chosen small enough, the derivative can be replaced
by the approximate expression
dy
y(t +
L
dt
t) - y(t)
t
(9.3–2)
Assume that the function f (t, y) in Equation (9.3–1) remains constant over the time
interval (t, t t), and replace Equation (9.3–1) by the following approximation:
y(t +
t) - y(t)
= f (t, y)
t
EULER METHOD
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or
y(t +
STEP SIZE
t) = y(t) + f (t, y) t
(9.3–3)
The smaller t is, the more accurate are our two assumptions leading to Equation (9.3–3). This technique for replacing a differential equation with a difference equation is the Euler method. The increment t is called the step size.
Equation (9.3–3) can be written in more convenient form as
y(tk + 1) = y(tk) +
t f [tk, y(tk)]
(9.3–4)
where tk1 tk t. This equation can be applied successively at the times tk by
putting it in a for loop. The accuracy of the Euler method can be improved sometimes by using a smaller step size. However, very small step sizes require longer
run times and can result in a large accumulated error due to roundoff effects.
The Predictor-Corrector Method
The Euler method can have a serious de ciency in problems where the variables
are rapidly changing, because the method assumes the variables are constant
over the time interval t. One way of improving the method is to use a better approximation to the right-hand side of Equation (9.3–1). Suppose instead of the
Euler approximation (9.3–4) we use the average of the right-hand side of Equation (9.3–1) on the interval (tk, tk1). This gives
y(tk + 1) = y(tk) +
t
( f + fk + 1)
2 k
(9.3–5)
where
fk = f [tk, y(tk)]
(9.3–6)
with a similar de nition for fk1. Equation (9.3–5) is equivalent to integrating
Equation (9.3–1) with the trapezoidal rule.
The dif culty with Equation (9.3–5) is that fk1 cannot be evaluated until
y(tk1) is known, but this is precisely the quantity being sought. A way out of this
dif culty is to use the Euler formula (9.3–4) to obtain a preliminary estimate of
y(tk1). This estimate is then used to compute fk1 for use in Equation (9.3–5) to
obtain the required value of y(tk1).
The notation can be changed to clarify the method. Let h t and yk y(tk),
and let xk1 be the estimate of y(tk1) obtained from the Euler formula (9.3–4).
Then, by omitting the tk notation from the other equations, we obtain the following description of the predictor-corrector process.
MODIFIED EULER
METHOD
Euler predictor
xk + 1 = yk + hf (tk, yk)
Trapezoidal corrector
yk + 1 = yk +
h
[ f (tk, yk) + f (tk + 1, xk + 1)]
2
(9.3–7)
(9.3–8)
This algorithm is sometimes called the modi ed Euler method . However, note
that any algorithm can be tried as a predictor or a corrector. Thus many other
methods can be classi ed as predictor -corrector.
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9.3
First-Order Differential Equations
Runge-Kutta Methods
The Taylor series representation forms the basis of several methods of solving
differential equations, including the Runge-Kutta methods. The Taylor series
may be used to represent the solution y(t h) in terms of y(t) and its derivatives
as follows.
1 $
#
(9.3–9)
y(t + h) = y(t) + hy(t) + h2 y(t) + . . .
2
The number of terms kept in the series determines its accuracy. The required
derivatives are calculated from the differential equation. If these derivatives can
be found, Equation (9.3–9) can be used to march forward in time. In practice, the
high-order derivatives can be dif cult to calculate, and the series (9.3–9) is truncated at some term. The Runge-Kutta methods were developed because of the
dif culty in computing the derivatives. These methods use several evaluations of
the function f (t, y) in a way that approximates the Taylor series. The number of
terms in the series that is duplicated determines the order of the Runge-Kutta
method. Thus, a fourth-order Runge-Kutta algorithm duplicates the Taylor series
through the term involving h4.
MATLAB ODE Solvers
In addition to the many variations of the predictor-corrector and Runge-Kutta algorithms that have been developed, there are more-advanced algorithms that use
a variable step size. These “adaptive” algorithms use larger step sizes when the
solution is changing more slowly. MATLAB provides several functions, called
solvers, that implement the Runge-Kutta and other methods with variable step
size. Two of these are the ode45 and ode15s functions. The ode45 function
uses a combination of fourth- and fth-order Runge-Kutta methods. It is a
general-purpose solver whereas ode15s is suitable for more-dif cult equations
called “stiff” equations. These solvers are more than suf cient to solve the problems in this text. It is recommended that you try ode45 rst. If the equation
proves dif cult to solve (as indicated by a lengthy solution time or by a warning
or error message), then use ode15s.
In this section we limit our coverage to rst-order equations. Solution of
higher-order equations is covered in Section 9.4. When used to solve the equa#
tion y = f (t, y), the basic syntax is (using ode45 as the example)
[t,y] = ode45(@ydot, tspan, y0)
where @ydot is the handle of the function le whose inputs must be t and y, and
whose output must be a column vector representing dy兾dt, that is, f (t, y). The
number of rows in this column vector must equal the order of the equation. The
syntax for ode15s is identical. The function le ydot may also be speci ed by
a character string (i.e., its name placed in single quotes), but use of the function
handle is now the preferred approach.
The vector tspan contains the starting and ending values of the independent variable t, and optionally any intermediate values of t where the solution is
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desired. For example, if no intermediate values are speci ed, tspan is [t0,
tfinal], where t0 and tfinal are the desired starting and ending values of
the independent parameter t. As another example, using tspan [0, 5, 10]
tells MATLAB to nd the solution at t 5 and at t 10. You can solve equation
backward in time by specifying t0 to be greater than tfinal.
The parameter y0 is the initial value y(0). The function le must have its rst
two input arguments as t and y in that order, even for equations where f (t, y) is not
a function of t. You need not use array operations in the function le because the
ODE solvers call the le with scalar values for the ar guments. The solvers may
have an additional argument, options, which is discussed at the end of this
section.
First consider an equation whose solution is known in closed form, so that
we can make sure we are using the method correctly.
Response of an RC Circuit
EXAMPLE 9.3–1
The model of the RC circuit shown in Figure 9.3–1 can be found from Kirchhoff’s
#
voltage law and conservation of charge. It is RCy + y = (t). Suppose the value of RC is
0.1 s. Use a numerical method to nd the free response for the case where the applied
voltage y is zero and the initial capacitor voltage is y(0) 2 V. Compare the results with
the analytical solution, which is y(t) 2e–10t.
Solution
#
#
The equation for the circuit becomes 0.1y + y = 0. First solve this for y: y = - 10y.
Next de ne and save the following function le. Note that the order of the input
arguments must be t and y even though t does not appear on the right-hand side of the
equation.
function ydot = RC_circuit(t,y)
% Model of an RC circuit with no applied voltage.
ydot = -10*y;
The initial time is t 0, so set t0 to be 0. Here we know from the analytical solution that
y(t) will be close to 0 for t 0.5 s, so we choose tfinal to be 0.5 s. In other problems
we generally do not have a good guess for tfinal, so we must try several increasing
values of tfinal until we see enough of the response on the plot.
R
+
v
C
y
–
Figure 9.3–1 An RC circuit.
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0.3
0.35
2
1.8
1.6
Capacitor Voltage
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
0.05
0.1
0.15
0.2
0.25
Time (s)
0.4
0.45
0.5
Figure 9.3–2 Free response of an RC circuit.
The function ode45 is called as follows, and the solution plotted along with the
analytical solution y_true.
[t, y] = ode45(@RC_circuit, [0, 0.5], 2);
y_true = 2*exp(-10*t);
plot(t,y,’o’,t,y_true), xlabel(‘Time(s)’),...
ylabel(‘Capacitor Voltage’)
Note that we need not generate the array t to evaluate y_true because t is generated
by the ode45 function. The plot is shown in Figure 9.3–2. The numerical solution is
marked by the circles, and the analytical solution is indicated by the solid line. Clearly the
numerical solution gives an accurate answer. Note that the step size has been automatically selected by the ode45 function.
Earlier versions of MATLAB required that the function name, here RC_
circuit, be enclosed within single quotes, but this might not be allowed in future
versions. The use of function handles is now preferred, such as @RC_circuit.
As we will see, additional capabilities are available with function handles.
Test Your Understanding
T9.3–1 Use MATLAB to compute and plot the solution of the following equation.
10
dy
+ y = 20 + 7 sin 2t
dt
y(0) = 15
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CHAPTER 9 Numerical Methods for Calculus and Differential Equations
When the differential equation is nonlinear, we often have no analytical solution to use for checking our numerical results. In such cases we can use our
physical insight to guard against grossly incorrect results. We can also check the
equation for singularities that might affect the numerical procedure. Finally, we
can sometimes use an approximation to replace the nonlinear equation with a linear one that can be solved analytically. Although the linear approximation does
not give the exact answer, it can be used to see if our numerical answer is “in the
ballpark.” The following example illustrates this approach.
Liquid Height in a Spherical Tank
EXAMPLE 9.3–2
Figure 9.3–3 shows a spherical tank for storing water. The tank is lled through a hole in the
top and drained through a hole in the bottom. If the tank’s radius is r, you can use integration to show that the volume of water in the tank as a function of its height h is given by
V(h) = rh2 - h3
3
(9.3–10)
Torricelli’s principle states that the liquid ow rate through the hole is proportional to the
square root of the height h. Further studies in uid mechanics have identi ed the relation
more precisely, and the result is that the volume ow rate through the hole is given by
(9.3–11)
q = Cd A12gh
where A is the area of the hole, g is the acceleration due to gravity, and Cd is an experimentally determined value that depends partly on the type of liquid. For water, Cd 0.6
is a common value. We can use the principle of conservation of mass to obtain a differential equation for the height h. Applied to this tank, the principle says that the rate of
change of liquid volume in the tank must equal the ow rate out of the tank; that is,
dV
= -q
dt
(9.3–12)
From Equation (9.3–10),
dh
dh
dh
dV
= 2rh
- h2
= h (2r - h)
dt
dt
dt
dt
r
h
Figure 9.3–3 Draining of
a spherical tank.
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9.4
Higher-Order Differential Equations
Substituting this and Equation (9.3–11) into Equation (9.3–12) gives the required equation for h.
(2rh - h2)
dh
= - Cd A12gh
dt
(9.3–13)
Use MATLAB to solve this equation to determine how long it will take for the tank
to empty if the initial height is 9 ft. The tank has a radius of r 5 ft and has a 1-in.-diameter
hole in the bottom. Use g 32.2 ft/sec2. Discuss how to check the solution.
Solution:
With Cd 0.6, r 5, g 32.2, and A (1兾24)2, Equation (9.3–13) becomes
0.03341 h
dh
= dt
10h - h2
(9.3–14)
We can rst check the above expression for dh兾dt for singularities. The denominator does
not become zero unless h 0 or h 10, which correspond to a completely empty and a
completely full tank. So we will avoid singularities if 0 h 10.
Finally, we can use the following approximation to estimate the time to empty. Replace
h on the right side of Equation (9.3–14) with its average value, namely, (9 0)兾2 4.5 ft.
This gives dh兾dt 0.00286, whose solution is h(t) h(0) 0.00286t 9 0.00286t.
According to this equation, the tank will be empty at t 9兾0.00286 3147 sec, or 52 min.
We will use this value as a “reality check” on our answer.
The function le based on Equation (9.3–14) is
function hdot height(t,h)
hdot = -(0.0334*sqrt(h))/(10*h-h^2);
The le is called as follows, using the ode45 solver.
[t, h]=ode45 (@height, [0, 2475], 9);
plot(t,h),xlabel(‘Time (sec)’), ylabel(‘Height (ft)’)
The resulting plot is shown in Figure 9.3–4. Note how the height changes more rapidly
when the tank is nearly full or nearly empty. This is to be expected because of the effects
of the tank’s curvature. The tank empties in 2475 sec, or 41 min. This value is not grossly
different from our rough estimate of 52 min, so we should feel comfortable accepting the
numerical results. The value of the nal time of 2475 sec was found by increasing the nal
time until the plot showed that the height became 0.
9.4 Higher-Order Differential Equations
To use the ODE solvers to solve an equation higher than order 1, you must rst
write the equation as a set of rst-order equations. This is easily done. Consider
the second-order equation
$
#
5y + 7y + 4y = f (t)
(9.4–1)
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9
8
7
Height (ft)
6
5
4
3
2
1
0
0
500
1000
Time (sec)
1500
2000
2500
Figure 9.3–4 Plot of water height in a spherical tank.
Solve it for the highest derivative:
4
7#
1
$
y = f (t) - y - y
5
5
5
(9.4–2)
#
De ne two new variables x1 and x2 to be y and its derivative y. That is, de ne
#
x1 y and x2 = y. This implies that
#
x1 = x2
1
4
7
#
x2 = f (t) - x1 - x2
5
5
5
CAUCHY OR
STATE-VARIABLE
FORM
This form is sometimes called the Cauchy form or the state-variable form.
#
#
Now write a function le that computes the values of x1 and x2 and stores
them in a column vector. To do this, we must rst have a function speci ed for
f (t). Suppose that f (t ) sin t. Then the required le is
function xdot = example_1(t,x)
% Computes derivatives of two equations
xdot(1) = x(2);
xdot(2) = (1/5)*(sin(t)–4*x(1)-7*x(2));
xdot = [xdot(1); xdot(2)];
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Higher-Order Differential Equations
391
#
#
Note that xdot(1) represents x1, xdot(2) represents x2, x(1) represents x1,
and x(2) represents x2. Once you become familiar with the notation for the
state-variable form, you will see that the previous code could be replaced with
the following shorter form.
function xdot = example_1(t,x)
% Computes derivatives of two equations
xdot = [x(2); (1/5)*(sin(t)-4*x(1)-7*x(2))];
Suppose we want to solve Equation (9.4–1) for 0 t 6 with the initial con#
ditions x(0) 3, x(0) = 9. Then the initial condition for the vector x is [3, 9].
To use ode45, you type
[t, x] = ode45(@example_1, [0, 6], [3, 9]);
Each row in the vector x corresponds to a time returned in the column vector t.
If you type plot(t,x), you will obtain a plot of both x1 and x2 versus t. Note
x is a matrix with two columns. The rst column contains the values of x1 at
the various times generated by the solver; the second column contains the values of x2. Thus, to plot only x1, type plot(t,x(:,1)). To plot only x2, type
plot(t,x(:,2)).
When we are solving nonlinear equations, sometimes it is possible to check
the numerical results by using an approximation that reduces the equation to a
linear one. The following example illustrates such an approach with a secondorder equation.
A Nonlinear Pendulum Model
EXAMPLE 9.4–1
The pendulum shown in Figure 9.4–1 consists of a concentrated mass m attached to a rod
whose mass is small compared to m. The rod’s length is L. The equation of motion for this
pendulum is
$
+
g
sin
L
(9.4–3)
= 0
Suppose that L 1 m and g 9.81 m/s2. Use MATLAB to solve this equation for (t) for
#
two cases: (0) 0.5 rad and (0) 0.8 rad. In both cases (0) = 0. Discuss how to
check the accuracy of the results.
Solution
If we use the small-angle approximation sin L , the equation becomes
$
+
g
L
(9.4–4)
= 0
which is linear and has the solution
(t) = (0) cos
g
t
L
(9.4–5)
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CHAPTER 9 Numerical Methods for Calculus and Differential Equations
L
θ
m
g
Figure 9.4–1 A pendulum.
#
if (0) = 0. Thus the amplitude of oscillation is (0), and the period is P 21L>g = 2.006 s. We can use this information to select a nal time and to check our
numerical results.
First rewrite the pendulum equation (9.4–3) as two rst-order equations. To do this,
#
let x1 = and x2 = . Thus
#
#
x1 = = x2
g
$
#
x2 = = - sin x1
L
The following function le is based on the last two equations. Remember that the
output xdot must be a column vector.
function xdot = pendulum(t,x)
g = 9.81; L = 1;
xdot = [x(2); –(g/L)*sin(x(1))];
This le is called as follows. The vectors ta and xa contain the results for the case
.
where (0) 0.5. In both cases, (0) = 0. The vectors tb and xb contain the results for
(0) 0.8.
[ta, xa] = ode45(@pendulum, [0,5], [0.5, 0];
[tb, xb] = ode45(@pendulum, [0,5], [0.8*pi, 0];
plot(ta, xa(:,1), tb,xb(:,1)), xlabel (‘Time (s)’), . . .
ylabel(‘Angle (rad)’), gtext(‘Case 1’), gtext(‘Case 2’)
The results are shown in Figure 9.4–2. The amplitude remains constant, as predicted
by the small-angle analysis, and the period for the case where (0) 0.5 is a little larger
than 2 s, the value predicted by the small-angle analysis. So we can place some con dence in the numerical procedure. For the case where (0) 0.8, the period of the
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Higher-Order Differential Equations
9.4
3
2
Case 2
Angle (rad)
1
0
Case 1
–1
–2
–3
0
0.5
1
1.5
2
2.5
Time (s)
3
3.5
4
4.5
5
Figure 9.4–2 The pendulum angle as a function of time for two starting positions.
numerical solution is about 3.3 s. This illustrates an important property of nonlinear
differential equations. The free response of a linear equation has the same period for any
initial conditions; however, the form and therefore the period of the free response of a
nonlinear equation often depend on the particular values of the initial conditions.
In this example, the values of g and L were encoded in the function
pendulum(t,x). Now suppose you want to obtain the pendulum response for
different lengths L or different gravitational accelerations g. You could use the
global command to declare g and L as global variables, or you could pass
parameter values through an argument list in the ode45 function; but starting
with MATLAB 7, the preferred method is to use a nested function. Nested functions are discussed in Section 3.3. The following program shows how this is done.
function pendula
g = 9.81; L = 0.75; % First case.
tF = 6*pi*sqrt(L/g); % Approximately 3 periods.
[t1, x1] = ode45(@pendulum, [0,tF], [0.4, 0];
%
g = 1.63; L = 2.5; % Second case.
tF = 6*pi*sqrt(L/g); % Approximately 3 periods.
[t2, x2] = ode45(@pendulum, [0,tF], [0.2, 0];
plot(t1, x1(:,1), t2, x2(:,1)), . . .
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CHAPTER 9 Numerical Methods for Calculus and Differential Equations
xlabel (‘time (s)’), ylabel (‘\theta (rad)’)
% Nested function.
function xdot = pendulum(t,x)
xdot = [x(2);–(g/L)*sin(x(1))];
end
end
Advanced Solver Capabilities
The complete preferred ODE solver syntax in MATLAB 7, using ode45 as an
example, is
[t, y] = ode45(@ydot, tspan, y0, options)
where the options argument is created with the odeset function.
The odeset Function The odeset function creates an options structure to
be supplied to the solver. Its syntax is
options = odeset(‘name1’, ‘value1’ ‘name2’, ‘value2’, . . .)
where name is the name of a property and value is the value to be assigned to
the property.
A simple example will clarify things. The Refine property is used to increase the number of output points from the solver by an integer factor n. For
ode45 the default value of n is 4 because of the solver’s large step sizes. Sup#
pose we want to solve the y = sin2 t for 0 … t … 4 with y(0) 0. De ne the
following function le.
function ydot = sinefn(t,y)
ydot = sin(t)^2;
Then use the odeset function to set the value of Refine to n 8, and call the
ode45 solver, as shown in the following code. This will produce twice as many
points to plot to obtain a smoother curve.
options = odeset(‘Refine’,8);
[t, y] = ode45(@sinefn, [0, 4*pi], 0, options);
Another property is the Events property, which has two possible values: on
and off. It can be used to locate transitions to, from, or through zeros of a userde ned function. This can be used to detect in the ODE solution when a variable
makes a transition to, from, or through a certain value, such as zero. This feature
can be used, for example, to simulate a dropped ball bouncing up from the oor .
See the MATLAB Help for other examples.
There are many properties that can be set with the odeset function. To see
a list of these, type odeset. Table 9.4–1 summarizes the syntax of the ODE
solvers using ode45 as an example.
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Special Methods for Linear Equations
395
Table 9.4–1 Syntax of the ODE solver ode45
Command
Description
[t, y] = ode45(@ydot,
tspan, y0, options)
#
Solves the vector differential equation y = f(t, y) speci ed by the function le
whose handle is @ydot and whose inputs must be t and y, and whose output
must be a column vector representing dy兾dt; that is, f(t, y). The number of rows
in this column vector must equal the order of the equation. The vector tspan
contains the starting and ending values of the independent variable t, and
optionally any intermediate values of t where the solution is desired. The
vector y0 contains the initial values. The function le must have two input
arguments, t and y, even for equations where f (t, y) is not a function of t.
The options argument is created with the odeset function. The syntax is
identical for the solver ode15s.
Creates an integrator options structure options to be used with the
ODE solver, in which the named properties have the speci ed values,
where name is the name of a property and value is the value to be
assigned to the property. Any unspeci ed properties have default values.
Typing odeset with no input arguments displays all property names and their
possible values.
options = odeset
(‘name1’, ‘value1’
‘name2’, ‘value2’,
. . .)
9.5 Special Methods for Linear Equations
MATLAB provides some convenient tools to use if the differential equation
model is linear. Even though there are general methods available for nding the
analytical solutions of linear differential equations, it is sometimes more convenient to use a numerical method to nd the solution. Examples of such situations
occur when the forcing function is a complicated function or when the order of
the differential equation is higher than 2. In such cases the algebra involved in
obtaining the analytical solution might not be worth the effort, especially if the
main objective is to obtain a plot of the solution.
Matrix Methods
We can use matrix operations to reduce the number of lines to be typed in the
derivative function le. For example, the following equation describes the motion of a mass connected to a spring, with viscous friction acting between the
mass and the surface. Another force u(t) also acts on the mass.
$
#
my + cy + ky = u(t)
(9.5–1)
#
This can be put into Cauchy form by letting x1 y and x2 = y. This gives
#
x1 = x2
1
k
c
#
x2 =
u(t) - x1 - x2
m
m
m
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CHAPTER 9 Numerical Methods for Calculus and Differential Equations
This can be written as one matrix equation as follows.
#
0
x1
c# d =
k
x2
Jm
In compact form this is
1
0
x1
d
c
+ 1 u(t)
c
- K x2
J K
m
m
#
x = Ax + Bu(t)
(9.5–2)
where
0
A = c-k
m
1
- mc d
0
B = c1 d
m
x
x = c 1d
x2
The following function le shows how to use matrix operations. In this example, m 1, c 2, k 5, and the applied force is u(t) 10.
function xdot = msd(t,x)
% Function le for mass with spring and damping.
% Position is rst variable, velocity is second variable.
u = 10;
m = 1;c = 2;k = 5;
A = [0, 1;-k/m, -c/m];
B = [0; 1/m];
xdot = A*x+B*u;
Note that the output xdot will be a column vector because of the de nition of
matrix-vector multiplication. We try different values of the nal time until we see
the entire response. Using a nal time of 5 and the initial conditions x1(0) 0
and x2(0) 0, we call the solver and plot the solution as follows:
[t, x] = ode45(@msd, [0,5], [0,0];
plot(t,x(:,1),t,x(:,2))
Figure 9.5–1 shows the edited plot. Note that we could have avoided embedding
the values of the parameters m, c, k, and u by making msd a nested function as
was done with the functions pendulum and pendula in Section 9.4.
Test Your Understanding
T9.5–1 Plot the position and velocity of a mass with a spring and damping, having the parameter values m 2, c 3, and k 7. The applied force is
u 35, the initial position is y(0) 2, and the initial velocity is
#
y(0) = - 3.
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Special Methods for Linear Equations
397
3
2.5
Displacement (m) and Velocity (m/s)
Displacement
2
1.5
1
Velocity
0.5
0
–0.5
–1
0
0.5
1
1.5
2
2.5
Time (s)
3
3.5
4
4.5
5
Figure 9.5–1 Displacement and velocity of the mass as a function of time.
Characteristic Roots from the eig Function
The characteristic roots of a linear differential equation give information about
the speed of response and the oscillation frequency, if any.
MATLAB provides the eig function to compute the characteristic roots
when the model is given in the state-variable form (9.5–2). Its syntax is eig(A),
where A is the matrix that appears in Equation (9.5–2). (The function’s name is
an abbreviation of eigenvalue, which is another name for characteristic root.) For
example, consider the equations
#
x1 = - 3x1 + x2
(9.5–3)
#
x2 = - x1 - 7x2
(9.5–4)
The matrix A for these equations is
A = c
-3
-1
1
d
-7
To nd the characteristic roots, type
>>A = [-3, 1;-1, -7];
>>r = eig(A)
The answer so obtained is r = [–6.7321, -3.2679]. To nd the time constants, which are the negative reciprocals of the real parts of the roots, you type
EIGENVALUE
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tau = -1./real(r). The time constants are 0.1485 and 0.3060. Four times
the dominant time constant, or 4(0.3060) 1.224, gives the time it takes for the
free response to become approximately zero.
ODE Solvers in the Control System Toolbox
LTI OBJECT
Many of the functions from the Control System toolbox are available in the
Student Edition of MATLAB. Some of these can be used to solve linear, timeinvariant (constant-coef cient) dif ferential equations. They are sometimes more
convenient to use and more powerful than the ODE solvers discussed thus far,
because general solutions can be found for linear, time-invariant equations. Here
we discuss several of these functions. These are summarized in Table 9.5–1. The
other features of the Control System toolbox require advanced methods, and will
not be covered here. See [Palm, 2005] for coverage of these methods.
An LTI object describes a linear, time-invariant equation, or sets of equations, here referred to as the system. An LTI object can be created from different
descriptions of the system, it can be analyzed with several functions, and it can
be accessed to provide alternate descriptions of the system. For example, the
equation
$
#
2x + 3x + 5x = u(t)
(9.5–5)
is one description of a particular system. This description is called the reduced
form. The following is a state-model description of the same system
#
x = Ax + Bu
(9.5–6)
#
where x1 x, x2 = x, and
A = c
0
- 52
1
- 32 d
0
B = c1 d
2
x
x = c 1d
x2
(9.5–7)
Both model forms contain the same information. However, each form has its
own advantages, depending on the purpose of the analysis.
Because there are two or more state variables in a state model, we need to be
able to specify which state variable, or combination of variables, constitutes the
output of the simulation. For example, models (9.5–6) and (9.5–7) can represent
the motion of a mass, with x1 the position and x2 the velocity of the mass. We need
to be able to specify whether we want to see a plot of the position, or the velocity,
or both. This speci cation of the output, denoted by the vector y, is done in general with the matrices C and D, which must be compatible with the equation
y = Cx + Du(t)
(9.5–8)
where the vector u(t) allows for multiple inputs. To continue the previous example, if we want the output to be the position x x1, then y x1, and we would
select C [1, 0] and D 0. Thus, in this case, Equation (9.5–8) reduces to y x1.
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To create an LTI object from the reduced form (9.5–5), use the tf(right,
left) function, and type
>>sys1 tf(1, [2, 3, 5]);
where the vector right is the vector of coef cients of the right-hand side of the
equation, arranged in descending derivative order, and left is the vector of
coef cients of the left-hand side of the equation, also arranged in descending
derivative order. The result, sys1, is the LTI object that describes the system in
reduced form, also called the transfer function form. (The name of the function,
tf, stands for transfer function, which is an equivalent way of describing the
coef cients on the left- and right-hand sides of the equation.)
The LTI object sys2 in transfer function form for the equation
6
d3x
d2x
dx
d2u
du
4
+
7
+
5x
=
3
+ 9
+ 2u
3
2
2
dt
dt
dt
dt
dt
(9.5–9)
is created by typing
>>sys2 = tf([3, 9, 2], [6, -4, 7, 5]);
To create an LTI object from a state model, you use the ss(A, B, C, D)
function, where ss stands for state space. For example, to create an LTI object
in state-model form for the system described by Equations (9.5–6) through
(9.5–8), you type
>>A = [0, 1; -5/2, -3/2]; B = [0; 1/2];
>>C = [1, 0]; D = 0;
>>sys3 = ss(A,B,C,D);
An LTI object de ned using the tf function can be used to obtain an equivalent state-model description of the system. To create a state model for the system described by the LTI object sys1 created previously in transfer function
form, you type ss(sys1). You will then see the resulting A, B, C, and D
matrices on the screen. To extract and save the matrices, use the ssdata
function as follows.
>>[A1, B1, C1, D1] = ssdata(sys1);
The results are
A1 = c
- 1.5
2
- 1.25
d
0
B1 = c
0.5
d
0
C1 = 30
0.54
D1 = [0]
When using ssdata to convert a transfer function form to a state model, note
that the output y will be a scalar that is identical to the solution variable of the
reduced form; in this case the solution variable of Equation (9.5–1) is the variable y. To interpret the state model, we need to relate its state variables x1 and x2
to y. The values of the matrices C1 and D1 tell us that the output variable
y 0.5x2. Thus we see that x2 2y. The other state variable x1 is related to x2 by
#
#
x2 = 2x1. Thus x1 = y.
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Table 9.5–1 LTI object functions
Command
Description
sys = ss(A, B, C, D)
Creates an LTI object in state-space form, where the matrices A, B,
.
C, and D correspond to those in the model x ⴝ Ax ⴙ Bu, y ⴝ
Cx ⴙ Du.
Extracts the matrices A, B, C and D corresponding to those in the
.
model x ⴝ Ax ⴙ Bu, y ⴝ Cx ⴙ Du.
Creates an LTI object in transfer function form, where the vector
right is the vector of coef cients of the right-hand side of the
equation, arranged in descending derivative order, and left is the
vector of coef cients of the left-hand side of the equation, also
arranged in descending derivative order.
Creates the transfer function model sys2 from the state model sys1.
Creates the state model sys1 from the transfer function model sys2.
Extracts the coef cients on the right- and left-hand sides of the
reduced-form model speci ed in the transfer function model sys.
When the optional parameter ‘v’ is used, the coef cients are
returned as vectors rather than as cell arrays.
[A, B, C, D] = ssdata(sys)
sys = tf(right,left)
sys2 = tf(sys1)
sys1 = ss(sys2)
[right, left] = tfdata(sys,’v’)
To create a transfer function description of the system sys3, previously
created from the state model, you type tfsys3 = tf(sys3). To extract and
save the coef cients of the reduced form, use the tfdata function as follows:
[right, left] = tfdata(sys3, ‘v’)
For this example, the vectors returned are right = 1 and left = [1, 1.5,
2.5]. The optional parameter ‘v’ tells MATLAB to return the coef cients as
vectors; otherwise, they are returned as cell arrays. These functions are summarized in Table 9.5–1.
Test Your Understanding
T9.5–2 Obtain the state model for the reduced-form model
$
#
5x + 7x + 4x = u(t)
Then convert the state model back to reduced form, and see if you get
the original reduced-form model.
Linear ODE Solvers
The Control System toolbox provides several solvers for linear models. These
solvers are categorized by the type of input function they can accept: zero input,
impulse input, step input, and a general input function. These are summarized in
Table 9.5–2.
The initial Function The initial function computes and plots the free
response of a state model. This is sometimes called the initial-condition response
or the undriven response in the MATLAB documentation. The basic syntax is
initial(sys,x0), where sys is the LTI object in state-model form and x0
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Table 9.5–2 Basic syntax of the LTI ODE solvers
Command
Description
impulse(sys)
Computes and plots the impulse response of the LTI object sys.
initial(sys,x0) Computes and plots the free response of the LTI object sys given in
state-model form, for the initial conditions speci ed in the vector x0.
lsim(sys,u,t)
Computes and plots the response of the LTI object sys to the input
speci ed by the vector u, at the times speci ed by the vector t.
step(sys)
Computes and plots the step response of the LTI object sys.
See the text for description of extended syntax.
is the initial-condition vector. The time span and number of solution points are
chosen automatically. For example, to nd the free response of the state model
(9.5–5) through (9.5–8), for x1(0) 5 and x2(0) –2, rst de ne it in statemodel form. This was done previously to obtain the system sys3. Then use the
initial function as follows.
>>initial(sys3, [5, -2])
The plot shown in Figure 9.5–2 will be displayed on the screen. Note that
MATLAB automatically labels the plot, computes the steady-state response, and
displays it with a dotted line.
Initial Condition Results
5
4
Amplitude
3
2
1
0
–1
0
1
2
3
4
5
6
7
Time (sec.)
Figure 9.5–2 Free response of the model given by Equations (9.5–5) through
(9.5–8) for x1(0) 5 and x2(0) –2.
8
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To specify the nal time tF, use the syntax initial(sys,x0,tF). To
specify a vector of times of the form t = 0:dt:tF, at which to obtain the solution, use the syntax initial(sys,x0,t).
When
called
with
left-hand
arguments,
as
[y, t, x] =
initial(sys,x0,. . .), the function returns the output response y, the
time vector t used for the simulation, and the state vector x evaluated at those
times. The columns of the matrices y and x are the outputs and the states,
respectively. The number of rows in y and x equals length(t). No plot is
drawn. The syntax initial(sys1, sys2, . . .,x0,t) plots the free
response of multiple LTI systems on a single plot. The time vector t is optional.
You can specify line color, line style, and marker for each system, for example,
initial(sys1,’r’, sys2,’y— —’,sys3,’gx’,x0).
The impulse Function The impulse function plots the unit-impulse response for each input-output pair of the system, assuming that the initial conditions are zero. (The unit impulse is also called the Dirac delta function.) The
basic syntax is impulse(sys), where sys is the LTI object. Unlike the
initial function, the impulse function can be used with either a state
model or a transfer function model. The time span and number of solution
points are chosen automatically. For example, the impulse response of Equation (9.5–5) is found as follows:
>>sys1 = tf(1, [2, 3, 5]);
>>impulse(sys1)
The extended syntax of the impulse function is similar to that of the initial
function.
The step Function The step function plots the unit-step response for each
input-output pair of the system, assuming that the initial conditions are zero.
[The unit step function u(t) is 0 for t 0 and 1 for t 0.] The basic syntax is
step(sys), where sys is the LTI object. The step function can be used with
either a state model or a transfer function model. The time span and number of
solution points are chosen automatically. The extended syntax of the step function is similar to that of the initial and the impulse functions.
To nd the unit-step response, for zero initial conditions, of the state model
(9.5–6) through (9.5–8), and the reduced-form model
#
$
#
5x + 7x + 5x = 5f + f
(9.5–10)
the session is (assuming sys3 is still available in the workspace)
>>sys4 = tf([5, 1], [5, 7, 5]);
>>step(sys3,’b’,sys4,’— —’)
The result is shown in Figure 9.5–3. The steady-state response is indicated by the
horizontal dotted line. Note how the steady-state response and the time to reach
that state are automatically determined.
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Step Response
0.7
0.6
Amplitude
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
8
Time (sec.)
Figure 9.5–3 Step response of the model given by Equations (9.5–6) through
(9.5–8) and the model (9.5–10), for zero initial conditions.
■
■
■
■
■
Step response can be characterized by the following parameters.
Steady-state value: The limit of the response as t : q .
Settling time: The time for the response to reach and stay within a certain
percentage (usually 2 percent) of its steady-state value.
Rise time: The time required for the response to rise from 10 to 90 percent
of its steady-state value.
Peak response: The largest value of the response.
Peak time: The time at which the peak response occurs.
When the step(sys) function puts a plot on the screen, you may use the plot
to calculate these parameters by right-clicking anywhere within the plot area.
This brings up a menu. Choose Characteristics to obtain a submenu that contains the response characteristics. When you select a speci c characteristic, for
example, “peak response,” MATLAB puts a large dot on the peak and displays
dashed lines indicating the value of the peak response and the peak time. Move
the cursor over this dot to see a display of the values. You can use the other solvers
in the same way, although the menu choices may be different. For example, peak
response and settling time are available when you use the impulse(sys)
function, but not the rise time. If instead of choosing Characteristics you choose
Properties and select the Options tab, you can change the defaults for the settling time and rise time, which are 2 percent and 10 to 90 percent.
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Using this method, we nd that the solid curve in Figure 9.5–3 has the
following characteristics:
■
■
■
■
■
Steady-state value: 0.2
2 percent settling time: 5.22
10 to 90 percent rise time: 1.01
Peak response: 0.237
Peak time: 2.26
You can also read values off any part of the curve by placing the cursor on
the curve at the desired point. You can move the cursor along the curve and read
the values as they change. Using this method, we nd that the solid curve in Figure 9.5–3 crosses the steady-state value of 0.2 for the second time at t 3.74.
You can suppress the plot generated by step and create your own plot as
follows, assuming sys3 is still available in the workspace.
[x,t] = step(sys3);
plot(t,x)
You can then use the Plot Editor tools to edit the plot. However, with this approach, right-clicking on the plot will no longer give you information about the
step response characteristics.
Suppose the step input is not a unit step but instead is 0 for t 0 and 10 for
t 0. There are two ways to obtain the solution with the factor 10. Using sys3
as the example, these are step(10*sys3)and
[x,t] = step(sys3);
plot(t,10*x)
The lsim Function The lsim function plots the response of the system to an
arbitrary input. The basic syntax for zero initial conditions is lsim(sys,u,t),
where sys is the LTI object, t is a time vector having regular spacing, as
t = 0:dt:tF, and u is a matrix with as many columns as inputs, and whose
ith row speci es the value of the input at time t(i). To specify nonzero initial
conditions for a state-space model, use the syntax lsim(sys,u,t,x0). This
computes and plots the total response (the free plus forced response). Rightclicking on the plot brings up the menu containing the Characteristics choice,
although the only characteristic available is the peak response.
When called with left-hand arguments, as [y, t] = lsim(sys, u,.. .),
the function returns the output response y and the time vector t used for the simulation. The columns of the matrix y are the outputs, and the number of its rows
equals length(t). No plot is drawn. To obtain the state vector solution for
state-space models, use the syntax [y, t, x] = lsim(sys,u,. . .).
The syntax lsim(sys1,sys2,. . .,u,t,x0) plots the responses of multiple LTI systems on a single plot. The initial-condition vector xo is needed
only if the initial conditions are nonzero. You can specify line color, line style,
and marker for each system, for example, lsim(sys1,’r’,sys2,
’y--’,sys3,’gx’,u,t).
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Special Methods for Linear Equations
405
L
cω
i
+
v
Keω
I
–
T = KT i
ω
Figure 9.5–4 An armature-controlled dc motor.
We will see an example of the lsim function shortly.
Programming Detailed Forcing Functions
As a nal example of higher -order equations, we now show how to program a
detailed forcing function for use with the lsim function. We use a dc motor as
the application. The equations for an armature-controlled dc motor (such as a
permanent-magnet motor) shown in Figure 9.5–4 are the following. They result
from Kirchhoff’s voltage law and Newton’s law applied to a rotating inertia. The
motor’s current is i and its rotational velocity is .
di
= - Ri - Ke + (t)
dt
d
= KT i - c
I
dt
L
(9.5–11)
(9.5–12)
where L, R, and I are the motor’s inductance, resistance, and inertia; KT and Ke
are the torque constant and back emf constant; c is a viscous damping constant;
and (t) is the applied voltage. These equations can be put into matrix form as
follows, where x1 i and x2 .
#
1
x
- R - KLe x1
L
d (t)
c # 1 d = c KLT
d
c
d
+
c
x2
- cI x2
0
I
Trapezoidal Pro le for a DC Motor
EXAMPLE 9.5–1
In many applications we want to accelerate the motor to a desired speed and allow it to
run at that speed for some time before decelerating to a stop. Investigate whether an applied
voltage having a trapezoidal pro le will accomplish this. Use the values R 0.6 , L 0.002 H, KT 0.04 N m/A, Ke 0.04 V s/rad, c 0, and I 6 10–5 kg m2. The
applied voltage in volts is given by
y(t) =
100t
10
μ
- 100(t - 0.4) + 10
0
0 … t 6 0.1
0.1 … t … 0.4
0.4 6 t … 0.5
t 7 0.5
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10
Voltage (V)
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6
4
2
0
0
0.1
0.2
0.3
t (s)
0.4
0.5
0.1
0.2
0.3
t (s)
0.4
0.5
300
Velocity (rad/s)
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200
100
0
0
0.6
Figure 9.5–5 Voltage input and resulting velocity response of a dc motor.
This is shown in the top graph in Figure 9.5–5.
■ Solution
The following program rst creates the model sys from the matrices A, B, C, and D. We
choose C and D to obtain the speed x2 as the only output. (To obtain both the speed and
the current as outputs, we would choose C = [1, 0; 0, 1] and D = [0; 0].) The
program computes the time constants using the eig function and then creates time, the
array of time values to be used by lsim. We choose the time increment 0.0001 to be a
very small fraction of the total time, 0.6 s.
The trapezoidal voltage function is then created with a for loop. This is perhaps the
easiest way because the if-elseif-else structure mimics the equations that de ne
(t). The initial conditions x1(0) and x2(0) are assumed to be zero, and so they need not be
speci ed in the lsim function.
% File dcmotor.m
R = 0.6; L = 0.002; c = 0;
K_T = 0.04; K_e = 0.04; I = 6e–5;
A = [-R/L, -K_e/L; K_T/I, -c/I];
B = [1/L; 0]; C = [0,1]; D = [0];
sys = ss(A,B,C,D);
Time_constants = -1./real(eig(A))
time = 0:0.0001:0.6;
k = 0;
for t = 0:0.0001:0.6
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k = k + 1;
if t < 0.1
v(k) = 100*t;
elseif t < = 0.4
v(k) = 10;
elseif t < = 0.5
v(k) = –100*(t-0.4) + 10;
else
v(k) = 0;
end
end
[y,t] = lsim(sys, v, time);
subplot(2,1,1), plot(time,v)
subplot(2,1,2), plot(time,y)
The time constants are computed to be 0.0041 and 0.0184 s. The largest time constant indicates that the motor’s response time is approximately 4(0.0184) 0.0736 s. Because this time is less than the time needed for the applied voltage to reach 10 V, the
motor should be able to follow the desired trapezoidal pro le reasonably well. To know
for certain, we must solve the motor’s differential equations. The results are plotted in the
bottom graph of Figure 9.5–5. The motor’s velocity follows a trapezoidal pro le as
expected, although there is some slight deviation because of its electric resistance and
mechanical inertia.
LTI Viewer The Control System toolbox contains the LTI Viewer, which
assists in the analysis of LTI systems. It provides an interactive user interface that
allows you to switch between different types of response plots and between the
analysis of different systems. The viewer is invoked by typing ltiview. See
the MATLAB Help for more information.
Prede ned Input Functions
You can always create any complicated input function to use with the ODE
solver ode45 or lsim by de ning a vector containing the input function’ s values at speci ed times, as was done in Example 9.5–1 for the trapezoidal pro le.
However, MATLAB provides the gensig function that makes it easy to construct periodic input functions.
The syntax [u, t] = gensig(type, period) generates a periodic
input of a speci ed type type, having a period period. The following types
are available: sine wave (type ‘sin’), square wave (type ‘square’), and
narrow-width periodic pulse (type ‘pulse’). The vector t contains the times,
and the vector u contains the input values at those times. All generated inputs
have unit amplitudes. The syntax [u, t] = gensig(type, period,
tF,dt) speci es the time duration tF of the input and the spacing dt between
the time instants.
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1.5
1
Response
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0.5
0
–0.5
0
1
2
3
4
5
Time
6
7
8
9
10
$
#
Figure 9.5–6 Square wave response of the model x + 2x + 4x = 4f.
For example, suppose a square wave with period 5 is applied to the following reduced-form model.
$
#
(9.5–13)
x + 2x + 4x = 4f
To nd the response for zero initial conditions, over the interval 0 t 10,
using a step size of 0.01, the session is
>>sys5 = tf(4,[1,2,4]);
>>[u, t] = gensig(‘square’,5,10,0.01);
>>[y, t] = lsim (sys5,u,t);plot(t,y,u), . . .
axis([0 10 -0.5 1.5]), . . .
xlabel(‘Time’),ylabel(‘Response’)
The result is shown in Figure 9.5–6.
9.6 Summary
This chapter covered numerical methods for computing integrals and derivatives, and for solving ordinary differential equations. Now that you have nished
this chapter, you should be able to do the following.
■
Numerically evaluate single, double, and triple integrals whose integrands
are given functions.
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■
■
■
■
■
■
■
■
■
Summary
Numerically evaluate single integrals whose integrands are given as
numerical values.
Numerically estimate the derivative of a set of data.
Compute the gradient and Laplacian of a given function.
Obtain in closed form the integral and derivative of a polynomial function.
Use the MATLAB ODE solvers to solve single rst-order ordinary
differential equations whose initial conditions are speci ed.
Convert higher-order ordinary differential equations into a set of rst-order
equations.
Use the MATLAB ODE solvers to solve sets of higher-order ordinary
differential equations whose initial conditions are speci ed.
Use MATLAB to convert a model from transfer function form to
state-variable form, and vice versa.
Use the MATLAB linear solvers to solve linear differential equations to obtain the free response and the step response for arbitrary forcing functions.
We have not covered all the differential equation solvers provided in
MATLAB, but limited our coverage to ordinary differential equations whose
initial conditions are speci ed. MA TLAB provides algorithms for solving boundaryvalue problems (BVPs) such as
$
#
x + 7x + 10x = 0
x(0) = 2
x(5) = 8
0 … t … 5
See the Help for the function bvp4c. Some differential equations are speci ed
#
implicitly as f(t, y, y) = 0. The solver ode15i can be used for such problems.
MATLAB can also solve delay-differential equations (DDEs) such as
$
#
x + 7x + 10x + 5x(t - 3) = 0
See the help for the functions dde23,
ddesd, and deval. The function
pdepe can solve partial differential equations. See also pdeval. In addition,
MATLAB provides support for analyzing and plotting the solver’s output. See
the functions odeplot, odephas2, odephas3, and odeprint.
Key Terms with Page References
Backward difference, 378
Cauchy form, 390
Central difference, 379
De nite integral, 370
Eigenvalue, 397
Euler method, 383
Forced response, 383
Forward difference, 378
Free response, 383
Improper integral, 370
Inde nite integral, 370
Initial-value problem (IVP), 382
Laplacian, 382
LTI object, 398
Modi ed Euler method, 384
Ordinary differential equation, 382
ODE, 382
Predictor-corrector method, 384
Quadrature, 373
Singularities, 370
State-variable form, 390
Step size, 384
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410
Problems
You can nd answers to problems marked with an asterisk at the end of the text.
Section 9.1
1.* An object moves at a velocity (t) 5 7t2 m/s starting from the position x(2) 5 m at t 2 s. Determine its position at t 10 s.
2.
The total distance traveled by an object moving at velocity (t) from the
time t a to the time t b is
b
x(b) =
La
ⱍ(t)ⱍ dt + x(a)
The absolute value ⱍ (t)ⱍ is used to account for the possibility that (t) might
be negative. Suppose an object starts at time t 0 and moves with a velocity
of (t) cos(t) m. Find the object’s location at t 1 s if x(0) 2 m.
3.
An object starts with an initial velocity of 3 m/s at t 0, and it
accelerates with an acceleration of a(t) 7t m/s2. Find the total distance
the object travels in 4 s.
4.
The equation for the voltage (t) across a capacitor as a function of time is
t
1
i(t) dt + Q0 d
(t) = c
C L0
where i(t) is the applied current and Q0 is the initial charge. A certain
capacitor initially holds no charge. Its capacitance is C 10–7 F. If a
current i(t) 0.2[1 sin (0.2t)] A is applied to the capacitor, compute
the voltage (t) at t 1.2 s if its initial velocity is zero.
5.
A certain object’s acceleration is given by a(t) 7t sin 5t m/s2. Compute
its velocity at t 10 s if its initial velocity is zero.
6.
A certain object moves with the velocity (t) given in the table below.
Determine the object’s position x(t) at t 10 s if x(0) 3.
Time (s)
0
1
2
3
4
5
6
7
8
9
10
Velocity (m/s)
0
2
5
7
9
12
15
18
22
20
17
7.* A tank having vertical sides and a bottom area of 100 ft2 is used to store
water. The tank is initially empty. To ll the tank, water is pumped into
the top at the rate given in the following table. Determine the water height
h(t) at t 10 min.
Time (min)
3
Flow rate (ft /min)
0
1
2
3
4
5
6
7
8
9
10
0
80
130
150
150
160
165
170
160
140
120
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Problems
8.
A cone-shaped paper drinking cup (like the kind supplied at water fountains) has a radius R and a height H. If the water height in the cup is h, the
water volume is given by
V =
R 2
1
a b h3
3
H
Suppose that the cup’s dimensions are R 1.5 in. and H 4 in.
a. If the ow rate from the fountain into the cup is 2 in. 3/s, how long will
it take to ll the cup to the brim?
b. If the ow rate from the fountain into the cup is given by 2(1 – e–2t)
in.3/s, how long will it take to ll the cup to the brim?
9.
A certain object has a mass of 100 kg and is acted on by a force
f (t) 500[2 – e–t sin(5t)] N. The mass is at rest at t 0. Determine the
object’s velocity at t 5 s.
10.* A rocket’s mass decreases as it burns fuel. The equation of motion for a
rocket in vertical ight can be obtained from Newton’ s law, and it is
m(t)
d
= T - m(t)g
dt
where T is the rocket’s thrust and its mass as a function of time is given
by m(t) m0(1 – rt兾b). The rocket’s initial mass is m0, the burn time is b,
and r is the fraction of the total mass accounted for by the fuel.
Use the values T 48,000 N, m0 2200 kg, r 0.8, g 9.81 m/s2, and
b 40 s. Determine the rocket’s velocity at burnout.
11.
The equation for the voltage (t) across a capacitor as a function of time is
t
y(t) =
1
c
i(t) dt + Q0 d
C L0
where i(t) is the applied current and Q0 is the initial charge. Suppose that
C 10–7 F and that Q0 0. Suppose the applied current is i(t) 0.3 0.1e–5t sin(25t) A. Plot the voltage (t) for 0 t 7 s.
12.
Compute the inde nite integral of p(x) 5x2 – 9x 8.
13.
Compute the double integral
3
A =
14.
3
L0 L1
(x2 + 3xy) dx dy
Compute the double integral
4
A =
L0 L0
x2 sin y dx dy
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CHAPTER 9 Numerical Methods for Calculus and Differential Equations
15.
Compute the double integral
1
A =
3
x2 (x + y) dx dy
L0 Ly
Note that the region of integration lies to the right of the line y x. Use
this fact and a MATLAB relational operator to eliminate values for which
y x.
16.
Compute the triple integral
2
A =
1
3
L1 L0 L1
xeyz dx dy dz
Section 9.2
17.
18.
Plot the estimate of the derivative dy兾dx from the following data. Do this
by using forward, backward, and central differences. Compare the results.
x
0
1
2
3
4
5
6
7
8
9
10
y
0
2
5
7
9
12
15
18
22
20
17
At a relative maximum of a curve y(x), the slope dy/dx is zero. Use the
following data to estimate the values of x and y that correspond to a maximum point.
x
0
1
2
3
4
5
6
7
8
9
10
y
0
2
5
7
9
10
8
7
6
8
10
19.
Compare the performance of the forward, backward, and central
difference methods for estimating the derivative of y(x) e–x sin(3x). Use
101 points from x 0 to x 4. Use a random additive error of 0.01.
20.
Compute the expressions for dp2兾dx, d( p1 p2)兾dx, and d( p2兾p1)兾dx for
p1 5x2 7 and p2 5x2 – 6x 7.
21.
Plot the contour plot and the gradient (shown by arrows) for the
function
f(x, y) = - x2 + 2xy + 3y2
Section 9.3
22.
Plot the solution of the equation
.
6y + y = f(t)
if f (t) 0 for t
y(0) 7.
0 and f(t) 15 for t
0. The initial condition is
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Problems
23.
The equation for the voltage y across the capacitor of an RC circuit is
RC
dy
+ y = (t)
dt
where (t) is the applied voltage. Suppose that RC 0.2 s and that the
capacitor voltage is initially 2 V. Suppose also that the applied voltage
goes from 0 to 10 V at t 0. Plot the voltage y(t) for 0 t 1 s.
24.
The following equation describes the temperature T(t) of a certain object
immersed in a liquid bath of constant temperature Tb.
10
dT
+ T = Tb
dt
Suppose the object’s temperature is initially T(0) 70F and the bath
temperature is Tb 170F.
a. How long will it take for the object’s temperature T to reach the bath
temperature?
b. How long will it take for the object’s temperature T to reach 168F?
c. Plot the object’s temperature T(t) as a function of time.
25.* The equation of motion of a rocket-propelled sled is, from Newton’s law,
.
m = f - c
where m is the sled mass, f is the rocket thrust, and c is an air resistance
coef cient. Suppose that m 1000 kg and c 500 N s/m. Suppose also
that (0) 0 and f 75,000 N for t 0. Determine the speed of the sled
at t 10 s.
26.
The following equation describes the motion of a mass connected to a
spring, with viscous friction on the surface.
$
#
my + cy + ky = 0
#
Plot y(t) for y(0) 10, y(0) = 5 if
a. m 3, c 18, and k 102
b. m 3, c 39 and k 120
27.
The equation for the voltage y across the capacitor of an RC circuit is
RC
dy
+ y = (t)
dt
where (t) is the applied voltage. Suppose that RC 0.2 s and that the
capacitor voltage is initially 2 V. Suppose also that the applied voltage is
(t) 10[2 – e–t sin(5t)] V. Plot the voltage y(t) for 0 t 5 s.
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CHAPTER 9 Numerical Methods for Calculus and Differential Equations
28.
The equation describing the water height h in a spherical tank with a drain
at the bottom is
(2rh - h2)
dh
= - Cd A 12gh
dt
Suppose the tank’s radius is r 3 m and the circular drain hole has a
radius of 2 cm. Assume that Cd 0.5 and that the initial water height is
h(0) 5 m. Use g 9.81 m/s2.
a. Use an approximation to estimate how long it takes for the tank to
empty.
b. Plot the water height as a function of time until h(t) 0.
29.
The following equation describes a certain dilution process, where y(t)
is the concentration of salt in a tank of freshwater to which salt brine is
being added.
dy
5
+
y = 4
dt
10 + 2t
Suppose that y(0) 0. Plot y(t) for 0 t 10.
Section 9.4
30.
The following equation describes the motion of a certain mass connected
to a spring, with viscous friction on the surface
$
#
3y + 18y + 102y = f(t)
where f (t) is an applied force. Suppose that f (t) 0 for t 0 and f (t) 10 for t 0.
#
a. Plot y(t) for y(0) = y(0) = 0.
#
b. Plot y(t) for y(0) 0 and y(0) = 10. Discuss the effect of the nonzero
initial velocity.
31.
32.
The following equation describes the motion of a certain mass connected
to a spring, with viscous friction on the surface
$
#
3y + 39y + 120y = f(t)
where f (t) is an applied force. Suppose that f(t) 0 for t 0 and f(t) 10
for t 0.
#
a. Plot y(t) for y(0) = y(0) = 0.
#
b. Plot y(t) for y(0) 0 and y(0) = 10. Discuss the effect of the nonzero
initial velocity.
The following equation describes the motion of a certain mass connected
to a spring, with no friction
$
3y + 75y = f(t)
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Problems
where f(t) is an applied force. Suppose the applied force is sinusoidal with
a frequency of rad/s and an amplitude of 10 N: f (t) 10 sin(t).
.
Suppose that the initial conditions are y(0) = y (0) = 0. Plot y(t) for 0 …
t … 20 s. Do this for the following three cases. Compare the results of
each case.
a. 1 rad/s
b. 5 rad/s
c. 10 rad/s
33.
Van der Pol’s equation has been used to describe many oscillatory
processes. It is
..
.
y - (1 - y2)y + y = 0
Plot y(t) for 1 and 0 t 20, using the initial conditions y(0) 5,
#
y(0) = 0.
34.
The equation of motion for a pendulum whose base is accelerating horizontally with an acceleration a(t) is
..
L + g sin = a(t) cos
.
Suppose that g 9.81 m/s2, L 1 m, and (0) = 0. Plot (t) for 0 t 10 s for the following three cases.
a. The acceleration is constant: a 5 m/s2, and (0) 0.5 rad.
b. The acceleration is constant: a 5 m/s2, and (0) 3 rad.
c. The acceleration is linear with time: a 0.5t m/s2, and (0) 3 rad.
35.
Van der Pol’s equation is
.
..
y - (1 - y2)y + y = 0
This equation is stiff for large values of the parameter . Compare the
performance of ode45 and ode15s for this equation. Use 1000
#
and 0 t 3000, with the initial conditions y(0) 2, y(0) = 0. Plot
y(t) versus t.
Section 9.5
36.
The equations for an armature-controlled dc motor are the following. The
motor’s current is i and its rotational velocity is .
di
= - Ri - Ke + (t)
(9.6–1)
dt
d
I
= KT i - c
(9.6–2)
dt
where L, R, and I are the motor’s inductance, resistance, and inertia; KT
and Ke are the torque constant and back emf constant; c is a viscous
damping constant; and (t) is the applied voltage.
L
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CHAPTER 9 Numerical Methods for Calculus and Differential Equations
Use the values R 0.8 , L 0.003 H, KT 0.05 N m/A, Ke 0.05 V s/rad, c 0, and I 8 10–5 kg m2.
a. Suppose the applied voltage is 20 V. Plot the motor’s speed and
current versus time. Choose a nal time lar ge enough to show the
motor’s speed becoming constant.
b. Suppose the applied voltage is trapezoidal as given below.
400t
20
(t) = μ
- 400(t - 0.2) + 20
0
0 … t 6 0.05
0.05 … t … 0.2
0.2 6 t … 0.25
t 7 0.25
Plot the motor’s speed versus time for 0 t 0.3 s. Also plot the applied
voltage versus time. How well does the motor speed follow a trapezoidal
pro le?
37.
Compute and plot the unit-impulse response of the following model.
$
#
10y + 3y + 7y = f(t)
38.
Compute and plot the unit-step response of the following model.
.
..
.
10y + 6y + 2y = f + 7f
39.* Find the reduced form of the following state model.
#
x
- 4 - 1 x1
2
c #1d = c
d c d + c d u(t)
2 - 3 x2
5
x2
40.
The following state model describes the motion of a certain mass connected to a spring, with viscous friction on the surface, where m 1,
c 2, and k 5.
#
x1
0
1 x1
0
c# d = c
d c d + c d f(t)
x2
- 5 - 2 x2
1
a. Use the initial function to plot the position x1 of the mass, if the
initial position is 5 and the initial velocity is 3.
b. Use the step function to plot the step response of the position and
velocity for zero initial conditions, where the magnitude of the step
input is 10. Compare your plot with that shown in Figure 9.5–1.
41.
Consider the following equation.
$
#
5y + 2y + 10y = f(t)
#
a. Plot the free response for the initial conditions y(0) 10, y(0) = - 5.
b. Plot the unit-step response (for zero initial conditions).
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Problems
c. The total response to a step input is the sum of the free response and
the step response. Demonstrate this fact for this equation by plotting
the sum of the solutions found in parts a and b and comparing the
plot with that generated by solving for the total response with
#
y(0) 10, y(0) = - 5.
42.
The model for the RC circuit shown in Figure P42 is
RC
do
+ o = i
dt
For RC 0.2 s, plot the voltage response o(t) for the case where the
applied voltage is a single square pulse of height 10 V and duration 0.4 s,
starting at t 0. The initial capacitor voltage is zero.
R
+
vi
–
Figure P42
C
vo
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Nick Koudis/Getty Images/RF
Engineering in the
21st Century. . .
Embedded Control Systems
A
n embedded control system is a microprocessor and sensor suite designed to be an integral part of a product. The aerospace and automotive
industries have used embedded controllers for some time, but the decreasing cost of components now makes embedded controllers feasible for more
consumer and biomedical applications.
For example, embedded controllers can greatly increase the performance of
orthopedic devices. One model of an arti cial leg now uses sensors to measure
in real time the walking speed, the knee joint angle, and the loading due to the
foot and ankle. These measurements are used by the controller to adjust the
hydraulic resistance of a piston to produce a more stable, natural, and ef cient
gait. The controller algorithms are adaptive in that they can be tuned to an individual’s characteristics and their settings changed to accommodate different
physical activities.
Engines incorporate embedded controllers to improve ef ciency . Embedded
controllers in new active suspensions use actuators to improve on the performance of traditional passive systems consisting only of springs and dampers.
One design phase of such systems is hardware-in-the-loop testing, in which the
controlled object (the engine or vehicle suspension) is replaced with a real-time
simulation of its behavior. This enables the embedded system hardware and software to be tested faster and less expensively than with the physical prototype,
and perhaps even before the prototype is available.
Simulink is often used to create the simulation model for hardware-in-the-loop
testing. The Control Systems and the Signal Processing toolboxes, and the DSP
and Fixed Point block sets, are also useful for such applications. ■
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C H A P T E R
10
Simulink
OUTLINE
10.1 Simulation Diagrams
10.2 Introduction to Simulink
10.3 Linear State-Variable Models
10.4 Piecewise-Linear Models
10.5 Transfer-Function Models
10.6 Nonlinear State-Variable Models
10.7 Subsystems
10.8 Dead Time in Models
10.9 Simulation of a Nonlinear Vehicle Suspension Model
10.10 Summary
Problems
Simulink is built on top of MATLAB, so you must have MATLAB to use
Simulink. It is included in the Student Edition of MATLAB and is also available
separately from The MathWorks, Inc. Simulink is widely used in industry to
model complex systems and processes that are dif cult to model with a simple
set of differential equations.
Simulink provides a graphical user interface that uses various types of elements called blocks to create a simulation of a dynamic system, that is, a system
that can be modeled with differential or difference equations whose independent
variable is time. For example, one block type is a multiplier, another performs a
sum, and still another is an integrator. The Simulink graphical interface enables
you to position the blocks, resize them, label them, specify block parameters, and
interconnect the blocks to describe complicated systems for simulation.
This chapter starts with simulations of simple systems that require few blocks.
Gradually, through a series of examples, more block types are introduced. The
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CHAPTER 10 Simulink
chosen applications require only a basic knowledge of physics and thus can be
appreciated by readers from any engineering or scienti c discipline. By the time
you have nished this chapter , you will have seen the block types needed to simulate a large variety of common applications.
10.1 Simulation Diagrams
BLOCK DIAGRAM
You develop Simulink models by constructing a diagram that shows the elements
of the problem to be solved. Such diagrams are called simulation diagrams or
.
block diagrams. Consider the equation y 10 f (t). Its solution can be represented
symbolically as
y(t) =
10 f (t) dt
L
which can be thought of as two steps, using an intermediate variable x:
x(t) dt
L
This solution can be represented graphically by the simulation diagram shown in
Figure 10.1–1a. The arrows represent the variables y, x, and f. The blocks represent
cause-and-effect processes. Thus, the block containing the number 10 represents
the process x(t)10f(t), where f(t) is the cause (the input) and x(t) represents the
effect (the output). This type of block is called a multiplier or gain block.
The block containing the integral sign 1 represents the integration process
y(t) = 1 x(t) dt, where x(t) is the cause (the input) and y(t) represents the effect
(the output). This type of block is called an integrator block.
There is some variation in the notation and symbols used in simulation diagrams. Figure 10.1–1b shows one variation. Instead of being represented by a
box, the multiplication process is now represented by a triangle like that used to
represent an electrical ampli er , hence the name gain block.
In addition, the integration symbol in the integrator block has been replaced
by the operator symbol 1/s, which derives from the notation used for the Laplace
transform (see Section 11.7 for a discussion of this transform). Thus the equation
y. = 10f(t) is represented by sy 10f, and the solution is represented as
x(t) = 10f (t)
GAIN BLOCK
INTEGRATOR
BLOCK
and
y =
y(t) =
10 f
s
or as the two equations
and
x = 10f
f (t)
10
x(t)
(a)
∫
y(t)
f
y =
10
1
x
s
x
(b)
.
Figure 10.1–1 Simulation diagrams for y = 10 f(t).
1
s
y
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10.2
x
z
x
z
f
1
s
y
y
Introduction to Simulink
421
y
10
(a)
(b)
Figure 10.1–2 (a) The summer element. (b) Simulation diagram for
y. = f (t) - 10y.
Another element used in simulation diagrams is the summer that, despite its
name, is used to subtract as well as to sum variables. Two versions of its symbol
are shown in Figure 10.1–2a. In each case the symbol represents the equation
z = x - y. Note that a plus or minus sign is required for each input arrow.
.
The summer symbol can be used to represent the equation y = f (t) - 10y,
which can be expressed as
y(t) =
L
or as
y =
SUMMER
[ f (t) - 10y] dt
1
( f - 10y)
s
You should study the simulation diagram shown in Figure 10.1–2b to con rm
that it represents this equation. This gure forms the basis for developing a
Simulink model to solve the equation.
10.2 Introduction to Simulink
Type simulink in the MATLAB Command window to start Simulink. The
Simulink Library Browser window opens. See Figure 10.2–1. The Simulink
blocks are located in “libraries.” These libraries are displayed under the Simulink
heading in Figure 10.2–1. Depending on what other MathWorks products are
installed, you might see additional items in this window, such as the Control
System Toolbox and State ow . These provide additional Simulink blocks, which
can be displayed by clicking on the plus sign to the left of the item. As Simulink
evolves through new versions, some libraries are renamed and some blocks are
moved to different libraries, so the library we specify here might change in later
releases. The best way to locate a block, given its name, is to type its name in the
search pane at the top of the Simulink Library Browser. When you press Enter,
Simulink will take you to the block location.
To create a new model, click on the icon that resembles a clean sheet of
paper, or select New from the File menu in the browser. A new Untitled window
opens for you to create the model. To select a block from the Library Browser,
double-click on the appropriate library, and a list of blocks within that library
then appears as shown in Figure 10.2–1. This gure shows the result of clicking
on the Continuous library, then clicking on the Integrator block.
LIBRARY
BROWSER
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CHAPTER 10 Simulink
Figure 10.2–1 The Simulink Library Browser.
Click on the block name or icon, hold the mouse button down, drag the block
to the new model window, and release the button. You can access help for that
block by right-clicking on its name or icon and selecting Help from the dropdown menu.
Simulink model les have the extension . mdl. Use the File menu in the
model window to Open, Close, and Save model les. To print the block diagram
of the model, select Print on the File menu. Use the Edit menu to copy, cut, and
paste blocks. You can also use the mouse for these operations. For example, to
delete a block, click on it and press the Delete key.
Getting started with Simulink is best done through examples, which we now
present.
#
Simulink Solution of y = 10 sin t
EXAMPLE 10.2–1
Use Simulink to solve the following problem for 0 … t … 13.
dy
= 10 sin t
dt
The exact solution is y (t) 10(1 – cos t).
y(0) = 0
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10.2
Introduction to Simulink
■ Solution
To construct the simulation, do the following steps. Refer to Figure 10.2–2. Figure 10.2–3
shows the Model window after completing the following steps.
1. Start Simulink and open a new model window as described previously.
2. Select and place in the new window the Sine Wave block from the Sources
library. Double-click on it to open the Block Parameters window, and make sure
the Amplitude is set to 1, the Bias to 0, the Frequency to 1, the Phase to 0, and the
Sample time to 0. Then click OK.
3. Select and place the Gain block from the Math Operations library, double-click on
it, and set the Gain value to 10 in the Block Parameters window. Then click OK.
Note that the value 10 then appears in the triangle. To make the number more
visible, click on the block, and drag one of the corners to expand the block so that
all the text is visible.
4. Select and place the Integrator block from the Continuous library, double-click
on it to obtain the Block Parameters window, and set the Initial condition to 0
[because y (0) 0]. Then click OK.
5. Select and place the Scope block from the Sinks library.
6. Once the blocks have been placed as shown in Figure 10.2–2, connect the input
port on each block to the outport port on the preceding block. To do this, move the
cursor to an input port or an output port; the cursor will change to a cross. Hold the
mouse button down, and drag the cursor to a port on another block. When you release
the mouse button, Simulink will connect them with an arrow pointing at the input
port. Your model should now look like that shown in Figure 10.2–2.
7. Enter 13 for the Stop time in the window to the right of the Start Simulation icon
(the black triangle). See Figure 10.2–3. The default value is 10, which can be
deleted and replaced with 13.
10
Sine Wave
1
s
Integrator
Gain
Scope
.
Figure 10.2–2 Simulink model for y = 10 sin t.
Figure 10.2–3 The Simulink Model window showing the model created in Example 10.2–1.
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Figure 10.2–4 The Scope window after running the
model in Example 10.2–1.
8.
9.
Run the simulation by clicking on the Start Simulation icon on the toolbar.
You will hear a bell sound when the simulation is nished. Then double-click on
the Scope block and click on the binoculars icon in the Scope display to enable
autoscaling. You should see an oscillating curve with an amplitude of 10 and a period
of 2 (Figure 10.2–4). The independent variable in the Scope block is time t; the
input to the block is the dependent variable y. This completes the simulation.
In the Con guration Parameters submenu under the Simulation menu,
you can select the ODE solver to use by clicking on the Solver tab. The default
is ode45, as indicated in the lower right-hand corner of the Model window.
To have Simulink automatically connect two blocks, select the Source block,
hold down the Ctrl key, and left-click on the Destination block. Simulink also provides easy ways to connect multiple blocks and lines; see the Help for information.
Note that blocks have a Block Parameters window that opens when you
double-click on the block. This window contains several items, the number and
nature of which depend on the speci c type of block. In general, you can use the
default values of these parameters, except where we have explicitly indicated
that they should be changed. You can always click on Help within the Block
Parameters window to obtain more information.
When you click on Apply, any changes immediately take effect and the window
remains open. If you click on OK, the changes take effect but the window closes.
Note that most blocks have default labels. You can edit text associated with
a block by clicking on the text and making the changes. You can save the
Simulink model as an .mdl le by selecting Save from the File menu in
Simulink. The model le can then be reloaded at a later time. You can also print
the diagram by selecting Print on the File menu.
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10.2
425
The Scope block is useful for examining the solution, but if you want to
obtain a labeled and printed plot, you can use the To Workspace block, which is
described in the next example.
Exporting to the MATLAB Workspace
EXAMPLE 10.2–2
We now demonstrate how to export the results of the simulation to the MATLAB
workspace, where they can be plotted or analyzed with any of the MATLAB functions.
■ Solution
Modify the Simulink model constructed in Example 10.2–1 as follows. Refer to Figure 10.2–5.
1. Delete the arrow connecting the Scope block by clicking on it and pressing the
Delete key. Delete the Scope block in the same way.
2. Select and place the To Workspace block from the Sinks library and the Clock
block from the Sources library.
3. Select and place the Mux block from the Signal Routing library, double-click on it,
and set the Number of inputs to 2. Click OK. (The name Mux is an abbreviation for
multiplexer, which is an electrical device for combining several signals.)
4. Connect the top input port of the Mux block to the output port of the Integrator
block. Then use the same technique to connect the bottom input port of the Mux
block to the outport port of the Clock block. Your model should now look like that
shown in Figure 10.2–5.
5. Double-click on the To Workspace block. You can specify any variable name you
want as the output; the default is simout. Change its name to y. The output variable y will have as many rows as there are simulation time steps, and as many
columns as there are inputs to the block. The second column in our simulation will
be time, because of the way we have connected the Clock to the second input port
of the Mux. Specify the Save format as Array. Use the default values for the other
parameters (these should be inf, 1, and -1 for Limit data points to last: Decimation, and Sample time, respectively). Click on OK.
10
Sine Wave
Gain
1
s
Integrator
y
Clock
To Workspace
Figure 10.2–5 Simulink model using the Clock and To
Workspace blocks.
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CHAPTER 10 Simulink
6.
After running the simulation, you can use the MATLAB plotting commands from
the Command window to plot the columns of y (or simout in general). To plot
y (t), type in the MATLAB Command window
>>plot(y(:,2),y(:,1)),xlabel(‘t’),ylabel(‘y’)
Simulink can be con gured to put the time variable tout into the MATLAB
workspace automatically when you are using the To Workspace block. This is
done with the Data Import/Export menu item under Con guration Parameters on the Simulation menu. The alternative is to use the Clock block to put
tout into the workspace. The Clock block has one parameter, Decimation. If
this parameter is set to 1, the Clock block will output the time every time step; if
set to 10, for example, the block will output every 10 time steps, and so on.
Simulink Model for y. = -10y + f (t)
EXAMPLE 10.2–3
Construct a Simulink model to solve
y. = -10y + f (t)
y (0) = 1
where f (t) = 2 sin 4 t, for 0 … t … 3.
■ Solution
To construct the simulation, do the following steps.
1. You can use the model shown in Figure 10.2–2 by rearranging the blocks as shown
in Figure 10.2–6. You will need to add a Sum block.
2. Select the Sum block from the Math Operations library and place it as shown in the
simulation diagram. Its default setting adds two input signals. To change this,
double-click on the block, and in the List of Signs window, type |+-. The signs
are ordered counterclockwise from the top. The symbol | is a spacer indicating
here that the top port is to be empty.
3. To reverse the direction of the Gain block, right-click on the block, select Format
from the pop-up menu, and select Flip Block.
4. When you connect the negative input port of the Sum block to the output port of
the Gain block, Simulink will attempt to draw the shortest line. To obtain the more
+–
Sine Wave
1
s
Integrator
Scope
10
Gain
Figure 10.2–6 Simulink model for y. = -10y + f(t).
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5.
6.
Linear State-Variable Models
427
standard appearance shown in Figure 10.2–6, rst extend the line vertically down
from the Sum input port. Release the mouse button, and then click on the end of the
line and attach it to the Gain block. The result will be a line with a right angle. Do
the same to connect the input of the Gain block to the arrow connecting the Integrator and the Scope blocks. A small dot appears to indicate that the lines have been
successfully connected. This point is called a takeoff point because it takes the
value of the variable represented by the arrow (here, the variable y) and makes that
value available to another block.
Set the Stop time to 3.
Run the simulation as before and observe the results in the Scope.
10.3 Linear State-Variable Models
State-variable models, unlike transfer-function models, can have more than one
input and more than one output. Simulink has the State-Space block that rep.
resents the linear state-variable model x ⴝAxⴙBu, y ⴝ Cx ⴙ Du. (See Section 9.5
for discussion of this model form.) The vector u represents the inputs, and the vector y represents the outputs. Thus, when you are connecting inputs to the StateSpace block, care must be taken to connect them in the proper order. Similar care
must be taken when connecting the block’s outputs to another block. The following example illustrates how this is done.
Simulink Model of a Two-Mass Suspension System
The following are the equations of motion of the two-mass suspension model shown in
Figure 10.3–1.
$
#
#
m1x 1 = k1(x2 - x1) + c1(x 2 - x1)
$
#
#
m2x 2 = - k1(x2 - x1) - c1(x2 - x1) + k2(y - x2)
Develop a Simulink model of this system to obtain the plots of x1 and x2. The input y(t) is
a unit step function, and the initial conditions are zero. Use the following values: m1 250 kg, m2 40 kg, k1 1.5 104 N/m, k2 1.5 105 N/m, and c1 1917 N # s/m.
■ Solution
The equations of motion can be expressed in state-variable form by letting z1 x1,
z2 x1, z3 x2, z4 x2. The equations of motion become
#
z1 = z2
1
#
z2 =
(- k1z1 - c1z2 + k1z3 + c1z4)
m1
#
z3 = z4
1
#
z4 =
[k1z1 + c1z2 - (k1 + k2)z3 - c1z4 + k2y]
m2
EXAMPLE 10.3–1
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CHAPTER 10 Simulink
Body
m1
x1
Suspension
k1
c1
m2
Wheel
x2
k2
Road
y
Datum level
Figure 10.3–1 Two-mass suspension model.
These equations are expressed in vector-matrix form as
#
z = Az + By(t)
where
0
- mk11
A = ≥
0
k1
m2
1
- mc11
0
0
0
k1
m1
c1
m1
c1
m2
- k1m+2k2
0
1 ¥
- mc12
0
0
B = ≥ ¥
0
k2
m2
and
z1
x1
#
z2
x1
z = ≥ ¥ = ≥ ¥
z3
x2
#
z4
x2
To simplify the notation, let a1 = k1>m1, a2 = c1>m1, a3 = k1>m2, a4 = c1>m2, a5 = k2>m2,
and a6 = a3 + a5. The matrices A and B become
0
-a1
A = ≥
0
a3
1
-a2
0
a4
0
a1
0
-a6
0
a2
1 ¥
-a4
0
0
B = ≥ ¥
0
a5
Next select appropriate values for the matrices in the output equation y Cz By(t).
Because we want to plot x1 and x2, which are z1 and z3, we must use the following
matrices for C and D.
C = c
1
0
0
0
0
1
0
d
0
0
D = c d
0
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10.3
Linear State-Variable Models
Note that the dimensions of B tell Simulink that there is one input. The dimensions of C
and D tell Simulink that there are two outputs.
Open a new model window, and then do the following to create the model shown in
Figure 10.3–2.
1. Select and place the Step block from the Sources library. Double-click on it to open
the Block Parameters window, and set the Step Time to 0, the Initial Value to 0, and
the Final Value to 1. Do not change the default value of any other parameters in this
window. Click OK. The Step Time is the time at which the step input begins.
2. Select and place the State-Space block from the Continuous library. Open its Block
Parameters window and enter the following values for the matrices A, B, C, and D.
For A enter
[0, 1, 0, 0; a1,a2, a1, a2; 0, 0, 0, 1; a3, a4, a6, a4]
For B enter [0; 0; 0; a5]. For C enter [1, 0, 0, 0; 0, 0, 1, 0], and
for D enter [0; 0]. Then enter [0; 0; 0; 0] for the initial conditions. Click OK.
x' = Ax+Bu
y = Cx+Du
Step
Scope
State-Space
Figure 10.3–2 Simulink model containing
the State-Space block and Step block.
1.4
x1
1.2
x2
Displacement (m)
1
0.8
0.6
0.4
0.2
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
Time (s)
Figure 10.3–3 Unit-step response of the two-mass suspension model.
0.9
1
429
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CHAPTER 10 Simulink
3. Select and place the Scope block from the Sinks library.
4. Connect the input and output ports as shown in Figure 10.3–2, and save the model.
5. In the Workspace window enter the parameter values and compute the ai constants as
shown in the following session.
>>m1 = 250; m2 = 40; k1 = 1.5e+4;
>>k2 = 1.5e+5; c1 =1917;
>>a1 = k1/m1; a2 = c1/m1; a3 = k1/m2;
>>a4 = c1/m2; a5 = k2/m2; a6 = a3 + a5;
6. Experiment with different values of Stop Time until the Scope shows that steady state
has been reached. Using this method a Stop Time of 1 s was found to be satisfactory.
The plots of both x1 and x2 will appear in the Scope. The default ode45 solver gave
jagged lines for the x2 plot, so the ode15s solver was used instead. A To Workspace
block can be added to obtain the plot in MATLAB. Figure 10.3–3 was created in
this way.
10.4 Piecewise-Linear Models
Unlike linear models, closed-form solutions are not available for most nonlinear
differential equations, and we must therefore solve such equations numerically.
A nonlinear ordinary differential equation can be recognized by the fact that the
dependent variable or its derivatives appear raised to a power or in a transcendental function. For example, the following equations are nonlinear.
#
$
#
#
y + 1y = 0
yy + 5y + y = 0
y + sin y = 0
Piecewise-linear models are actually nonlinear, although they may appear to
be linear. They are composed of linear models that take effect when certain
conditions are satis ed. The effect of switching back and forth between these
linear models makes the overall model nonlinear. An example of such a model is
a mass attached to a spring and sliding on a horizontal surface with Coulomb
friction. The model is
#
f(t) - mg
if x Ú 0
$
#
mx + kx = e
f(t) + mg
if x 6 0
These two linear equations can be expressed as the single, nonlinear equation
#
+1
if x Ú 0
$
#
#
mx + kx = f(t) - mg sign(x )
where
sign(x ) = e
#
-1
if x 6 0
Solutions of models that contain piecewise-linear functions are very tedious
to program. However, Simulink has built-in blocks that represent many of the
commonly found functions such as Coulomb friction. Therefore Simulink is
especially useful for such applications. One such block is the Saturation block in
the Discontinuities library. The block implements the saturation function shown
in Figure 10.4–1.
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10.4
Piecewise-Linear Models
431
Output
Upper Limit
Slope 1
Input
Lower Limit
Figure 10.4–1 The saturation nonlinearity.
Simulink Model of a Rocket-Propelled Sled
EXAMPLE 10.4–1
A rocket-propelled sled on a track is represented in Figure 10.4–2 as a mass m with an applied force f that represents the rocket thrust. The rocket thrust initially is horizontal, but
the
$ engine accidentally pivots during ring and rotates with an angular acceleration of
= >50 rad/s. Compute the sled’s velocity ␷ for 0 … t … 6 if ␷(0) = 0. The rocket
thrust is 4000 N and the sled mass is 450 kg.
The sled’s equation of motion is
#
450␷ = 4000 cos (t)
To obtain (t), note that
t
#
=
L0
##
dt =
t
50
v
m
f
Figure 10.4–2 A rocket-propelled sled.
␪
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CHAPTER 10 Simulink
and
t
.
=
t
2
t dt =
t
100
L0 50
dt =
L0
Thus the equation of motion becomes
2
.
t b
450␷ = 4000 cos a
100
or
2
. 80
␷=
cos a
t b
9
100
The solution is formally given by
t
␷(t) =
80
2
t b dt
cos a
9 L0
100
Unfortunately, no closed-form solution is available for the integral, which is called
Fresnel’s cosine integral. The value of the integral has been tabulated numerically, but we
will use Simulink to obtain the solution.
(a) Create a Simulink model to solve this problem for 0 … t … 10 s.
(b) Now suppose that the engine angle is limited by a mechanical stop to 60 , which is
60/180 rad. Create a Simulink model to solve the problem.
■ Solution
..(a) There are several ways to create the input function
= /50 rad/s and that
= (/100)t2. Here we note that
t
.
..
=
L0
dt
and
t
.
2
t
100
..
Thus we can create (t) by integrating the constant = /50 twice. The simulation diagram is shown in Figure 10.4–3. This diagram is used to create the corresponding
Simulink model shown in Figure 10.4–4.
There are two new blocks in this model. The Constant block is in the Sources
library. After placing it, double-click on it and type pi/50 in its Constant Value
window.
The Trigonometric block is in the Math Operations library. After placing it, doubleclick on it and select cos in its Function window.
=
¨
1
s
˙
1
s
L0
dt =
cos
80
9
v
1
s
v
Figure 10.4–3 Simulation diagram for ␷ = (80/ 9) cos (t 2/100).
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10.4
pi/50
1
s
Constant
Integrator
1
s
cos
Piecewise-Linear Models
80/9
1
s
Gain
Integrator 2
Integrator 1 Trigonometric
Function
433
Scope
Figure 10.4–4 Simulink model for ␷ = (80/ 9) cos (t 2/100).
1
s
Constant 1 Integrator 3
80/9
1
1
cos
80/9
s
s
Constant Integrator Integrator 1 Saturation Trigonometric Gain
Function
pi/50
1
s
Integrator 2
Figure 10.4–5 Simulink model for ␷ = (80/ 9) cos (t 2/100) with a Saturation block.
Set the Stop Time to 10, run the simulation, and examine the results in the Scope.
(b) Modify the model in Figure 10.4–4 as follows to obtain the model shown in Figure 10.4–5. We use the Saturation block in the Discontinuities library to limit the range
of to 60/180 rad. After placing the block as shown in Figure 10.4–5, double-click
on it and type 60*pi/180 in its Upper Limit window. Then type 0 in its Lower Limit
window.
Enter and connect the remaining elements as shown, and run the simulation. The
upper Constant block and Integrator block are used to generate the solution when the engine angle is = 0, as a check on our results. [The equation of motion for = 0 is
.
␷ = 80>9, which gives ␷(t) = 80t>9.]
If you prefer, you can substitute a To Workspace block for the Scope. Then you
can plot the results in MATLAB. The resulting plot is shown in Figure 10.4–6.
The Relay Block
The Simulink Relay block is an example of something that is tedious to program
in MATLAB but is easy to implement in Simulink. Figure 10.4–7a is a graph of
the logic of a relay. The relay switches the output between two speci ed values,
named On and Off in the gure. Simulink calls these values “Output when on”
and “Output when off.” When the relay output is On, it remains On until the
input drops below the value of the Switch-off point parameter, named SwOff in
the gure. When the relay output is Off, it remains Off until the input exceeds the
value of the Switch-on point parameter, named SwOn in the gure.
The Switch-on point parameter value must be greater than or equal to the
Switch-off point value. Note that the value of Off need not be zero. Note also that
the value of Off need not be less than the value of On. The case where Off > On is
Scope
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CHAPTER 10 Simulink
60
=0
50
v (t ) (m/s)
40
≠0
30
20
10
0
0
1
2
3
t (s)
Figure 10.4–6 Speed response of the sled for
On
4
5
= 0 and
0.
6
Off
Off
On
SwOff
SwOn
SwOff
(a)
SwOn
(b)
Figure 10.4–7 The relay function. (a) The case where
On > Off. (b) The case where On < Off.
shown in Figure 10.4–7b. As we will see in the following example, it is sometimes necessary to use this case.
Model of a Relay-Controlled Motor
EXAMPLE 10.4–2
The model of an armature-controlled dc motor was discussed in Section 9.5. See Figure 10.4–8. The model is
L
di
= -Ri - Ke
dt
+ ␷(t)
I
d
= KT i - c
dt
- Td (t)
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10.4
R
Piecewise-Linear Models
L
cω
i
+
v
I
Ke ω
–
ω
T = KT i
Figure 10.4–8 An armature-controlled dc motor.
where the model now includes a torque Td (t ) acting on the motor shaft due, for example,
to some unwanted source such as Coulomb friction or wind gusts. Control system engineers call this a disturbance. These equations can be put into matrix form as follows,
where x1 = i and x2 = .
R
.
x1
L
c. d = ≥
KT
x2
I
Ke
1
x1
L
L
¥
+ ≥
c c x2 d
0
I
-
0
1
I
¥ c
␷ (t)
Td (t ) d
Use the values R = 0.6 , L = 0.002 H, K T = 0.04 N # m/A, Ke = 0.04 V # s/rad,
c = 0.01 N # m # s/rad, and I = 6 * 10 - 5 kg # m2.
Suppose we have a sensor that measures the motor speed, and we use the sensor’s
signal to activate a relay to switch the applied voltage ␷(t) between 0 and 100 V to keep
the speed between 250 and 350 rad/s. This corresponds to the relay logic shown in
Figure 10.4–7b, with SwOff 250, SwOn 350, Off 100, and On 0. Investigate
how well this scheme will work if the disturbance torque is a step function that increases
from 0 to 3 N # m, starting at t 0.05 s. Assume that the system starts from rest with
(0) = 0 and i(0) = 0.
■ Solution
For the given parameter values,
A = c
-300
666.7
- 20
d
- 166.7
B = c
500
0
0
d
-16 667
To examine the speed as output, we choose C [0, 1] and D [0, 0]. To create this
simulation, rst obtain a new model window . Then do the following.
1. Select and place in the new window the Step block from the Sources library. Label
it Disturbance Step as shown in Figure 10.4–9. Double-click on it to obtain the
Block Parameters window, and set the Step Time to 0.05, the Initial and Final values
to 0 and 3, and the Sample time to 0. Click OK.
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CHAPTER 10 Simulink
Relay
signal1
signal2
Disturbance
Step
Mux
x' = Ax + Bu
y = Cx + Du
State-Space
Scope
Figure 10.4–9 Simulink model of a relay-controlled motor.
2. Select and place the Relay block from the Discontinuities library. Double-click on
it, and set the Switch-on and Switch-off points to 350 and 250, and set the Output
when on and Output when off to 0 and 100. Click OK.
3. Select and place the Mux block from the Signal Routing library. The Mux block
combines two or more signals into a vector signal. Double-click on it, and set the
Display option to signals. Click OK. Then click on the Mux icon in the model window, and drag one of the corners to expand the box so that all the text is visible.
4. Select and place the State-Space block from the Continuous library. Double-click on
it, and enter [-300, -20; 666.7, -166.7] for A, [500, 0; 0, -16667]
for B, [0, 1] for C, and [0, 0] for D. Then enter [0; 0] for the initial conditions. Click OK. Note that the dimension of the matrix B tells Simulink that there
are two inputs. The dimensions of the matrices C and D tell Simulink that there is
one output.
5. Select and place the Scope block from the Sinks library.
6. Once the blocks have been placed, connect the input port on each block to the
outport port on the preceding block as shown in the gure. It is important to connect
the top port of the Mux block [which corresponds to the rst input, ␷ (t)] to the
output of the Relay block, and to connect the bottom port of the Mux block [(which
corresponds to the second input, Td (t)] to the output of the Disturbance Step block.
7. Set the Stop time to 0.1 (which is simply an estimate of how long is needed to
see the complete response), run the simulation, and examine the plot of (t) in the
Scope. You should see something like Figure 10.4–10. If you want to examine the
current i(t), change the matrix C to [1, 0] and run the simulation again.
The results show that the relay logic control scheme keeps the speed within the desired
limits of 250 and 350 before the disturbance torque starts to act. The speed oscillates because when the applied voltage is zero, the speed decreases as a result of the back-emf
and the viscous damping. The speed drops below 250 when the disturbance torque starts
to act, because the applied voltage is 0 at that time. As soon as the speed drops below 250,
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Transfer-Function Models
437
Figure 10.4–10 Scope display of the speed response of a
relay-controlled motor.
the relay controller switches the voltage to 100, but it now takes longer for the speed to
increase because the motor torque must now work against the disturbance.
Note that the speed becomes constant, instead of oscillating. This is so because with
␷ = 100, the system achieves a steady-state condition in which the motor torque equals
the sum of the disturbance torque and the viscous damping torque. Thus the acceleration
is zero.
One practical use of this simulation is to determine how long the speed is below the
limit of 250. The simulation shows that this time is approximately 0.013 s. Other uses of
the simulation include nding the period of the speed’ s oscillation (about 0.013 s) and the
maximum value of the disturbance torque that can be tolerated by the relay controller (it
is about 3.7 N m).
#
10.5 Transfer-Function Models
The equation of motion of a mass-spring-damper system is
..
.
my + cy + ky = f (t)
(10.5–1)
As with the Control System toolbox, Simulink can accept a system description in
transfer-function form and in state-variable form. (See Section 9.5 for a discussion of these forms.) If the mass-spring system is subjected to a sinusoidal forcing function f (t), it is easy to use the MATLAB commands presented thus far to
solve and plot the response y (t). However, suppose that the force f (t) is created
by applying a sinusoidal input voltage to a hydraulic piston that has a dead-zone
nonlinearity. This means that the piston does not generate a force until the input
DEAD ZONE
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0.5
0.4
0.3
Output
0.2
0.1
0
–0.1
–0.2
–0.3
–0.4
–0.5
–1 –0.8 –0.6 –0.4 –0.2
0 0.2 0.4 0.6 0.8
Input
1
Figure 10.5–1 A dead-zone nonlinearity
voltage exceeds a certain magnitude, and thus the system model is piecewise
linear.
A graph of a particular dead-zone nonlinearity is shown in Figure 10.5–1.
When the input (the independent variable on the graph) is between 0.5 and 0.5,
the output is zero. When the input is greater than or equal to the upper limit of
0.5, the output is the input minus the upper limit. When the input is less than or
equal to the lower limit of 0.5, the output is the input minus the lower limit. In
this example, the dead zone is symmetric about 0, but it need not be in general.
Simulations with dead-zone nonlinearities are somewhat tedious to program
in MATLAB, but are easily done in Simulink. The following example illustrates
how it is done.
Response with a Dead Zone
EXAMPLE 10.5–1
Create and run a Simulink simulation of a mass-spring-damper model (Equation 10.5–1)
using the parameter values m 1, c 2, and k 4. The forcing function is the function
f (t) sin 1.4t. The system has the dead-zone nonlinearity shown in Figure 10.5–1.
■ Solution
To construct the simulation, do the following steps.
1. Start Simulink and open a new Model window as described previously.
2. Select and place in the new window the Sine Wave block from the Sources library.
Double-click on it, and set the Amplitude to 1, the Frequency to 1.4, the Phase to 0,
and the Sample time to 0. Click OK.
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Transfer-Function Models
3. Select and place the Dead Zone block from the Discontinuities library, double-click
on it, and set the Start of dead zone to 0.5 and the End of dead zone to 0.5.
Click OK.
4. Select and place the Transfer Fcn block from the Continuous library, double-click on
it, and set the Numerator to [1] and the Denominator to [1, 2, 4]. Click OK.
5. Select and place the Scope block from the Sinks library.
6. Once the blocks have been placed, connect the input port on each block to the outport port on the preceding block. Your model should now look like Figure 10.5–2.
7. Set the Stop time to 10.
8. Run the simulation. You should see an oscillating curve in the Scope display.
It is informative to plot both the input and the output of the Transfer Fcn block versus
time on the same graph. To do this:
1. Delete the arrow connecting the Scope block to the Transfer Fcn block. Do this by
clicking on the arrow line and then pressing the Delete key.
2. Select and place the Mux block from the Signal Routing library, double-click on it,
and set the Number of inputs to 2. Click OK.
3. Connect the top input port of the Mux block to the output port of the Transfer Fcn
block. Then use the same technique to connect the bottom input port of the Mux
block to the arrow from the outport port of the Dead Zone block. Just remember to
start with the input port. Simulink will sense the arrow automatically and make the
connection. Your model should now look like Figure 10.5–3.
4. Set the Stop time to 10, run the simulation as before, and bring up the Scope display.
You should see what is shown in Figure 10.5–4. This plot shows the effect of the
dead zone on the sine wave.
1
s2+2s+4
Sine Wave
Dead Zone Transfer Fcn
Scope
Figure 10.5–2 The Simulink model of dead-zone
response.
Sine Wave Dead Zone
1
s2+2s+4
Transfer Fcn
Scope
Figure 10.5–3 Modi cation of the dead-zone model to
include a Mux block.
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Figure 10.5–4 The response of the dead-zone model.
Sine Wave
Dead Zone
1
s2+2s+4
Transfer Fcn 1
1
s2+2s+4
Transfer Fcn 2
Mux 1
Scope
simout
Clock
Mux 2 To Workspace
Figure 10.5–5 Modi cation of the dead-zone model to export variables to the
MATLAB workspace.
You can bring the simulation results into the MATLAB workspace by using the To
Workspace block. For example, suppose we want to examine the effects of the dead zone
by comparing the response of the system with and without a dead zone. We can do this
with the model shown in Figure 10.5–5. To create this model:
1. Copy the Transfer Fcn block by right-clicking on it, holding down the mouse button,
and dragging the block copy to a new location. Then release the button. Copy the
Mux block in the same way.
2. Double-click on the rst Mux block and change the number of its inputs to 3.
3. In the usual way, select and place the To Workspace block from the Sinks library
and optionally the Clock box from the Sources library. Double-click on the To
Workspace block. You can specify any variable name you want as the output; the
default is simout. Change its name to y. The output variable y will have as many
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441
rows as there are simulation time steps, and as many columns as there are inputs to
the block. The fourth column in our simulation will be time, because of the way we
have connected the Clock to the second Mux. Specify the Save format as Matrix. Use
the default values for the other parameters (these should be inf, 1, and -1 for Maximum number of rows, Decimation, and Sample time, respectively). Click on OK.
4. Connect the blocks as shown, and run the simulation.
5. You can use the MATLAB plotting commands from the Command window to plot
the columns of y; for example, to plot the response of the two systems and the
output of the Dead Zone block versus time, type
>>plot(y(:,4),y(:,1),y(:,4),y(:,2),y(:,4),y(:,3))
10.6 Nonlinear State-Variable Models
Nonlinear models cannot be put into transfer-function form or the state-variable
#
form x ⴝ Ax ⴙ Bu. However, they can be simulated in Simulink. The following
example shows how this can be done.
Model of a Nonlinear Pendulum
EXAMPLE 10.6–1
The pendulum shown in Figure 10.6–1 has the following nonlinear equation of motion,
if there is viscous friction in the pivot and if there is an applied moment M(t) about the
pivot
$
#
I + c + mgL sin = M(t)
where I is the mass moment of inertia about the pivot. Create a Simulink model for this
system for the case where I 4, mgL 10, c 0.8, and M(t) is a square wave with an
amplitude of 3 and # a frequency of 0.5 Hz. Assume that the initial conditions are
(0) = >4 rad and (0) = 0.
■ Solution
To simulate this model in Simulink, de ne a set of # variables that lets you rewrite the
equation as two rst-order equations. Thus let = . Then the model can be written as
#
=
#
=
1
[- c
I
- mgL sin
+ M(t)] = 0.25[- 0.8
- 10 sin
Integrate both sides of each equation over time to obtain
=
L
= 0.25
dt
L
[- 0.8
- 10 sin
+ M(t)] dt
+ M(t)]
L
␪
m
g
Figure 10.6–1
A pendulum.
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CHAPTER 10 Simulink
Signal
Generator
–+
–
1
1
s
s
Integrator 1 Integrator 2
0.25
1/I
Scope
0.8
c
10*sin(u)
Fcn
Figure 10.6–2 Simulink model of nonlinear pendulum dynamics.
We will introduce four new blocks to create this simulation. Obtain a new Model window
and do the following.
1. Select and place in the new window the Integrator block from the Continuous
library, and change its label to Integrator 1 as shown in Figure 10.6–2. You can
edit text associated with a block by clicking on the text and making the changes.
Double-click on the block to obtain the Block Parameters
window, and set the
#
Initial condition to 0 [this is the initial condition (0) = 0]. Click OK.
2. Copy the Integrator block to the location shown and change its label to Integrator 2.
Set its initial condition to >4 by typing pi/4 in the Block Parameters window.
This is the initial condition (0) = >4.
3. Select and place a Gain block from the Math Operations library, double-click on it,
and set the Gain value to 0.25. Click OK. Change its label to 1/I. Then click on the
block, and drag one of the corners to expand the box so that all the text is visible.
4. Copy the Gain box, change its label to c, and place it as shown in Figure 10.6–2.
Double-click on it, and set the Gain value to 0.8. Click OK. To ip the box left to
right, right-click on it, select Format, and select Flip.
5. Select and place the Scope block from the Sinks library.
6. For the term 10 sin , we cannot use the Trig function block in the Math library
because we need to multiply the sin by 10. So we use the Fcn block under the
User-De ned Functions library (Fcn stands for function). Select and place this
block as shown. Double-click on it, and type 10*sin(u) in the expression window. This block uses the variable u to represent the input to the block. Click OK.
Then ip the block.
7. Select and place the Sum block from the Math Operations library. Double-click
on it, and select round for the Icon shape. In the List of Signs window, type +--.
Click OK.
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Subsystems
8. Select and place the Signal Generator block from the Sources library. Double-click
on it, select square wave for the Wave form, 3 for the Amplitude, 0.5 for the
Frequency, and Hertz for the Units. Click OK.
9. Once the blocks have been placed, connect arrows as shown in the gure.
10. Set the Stop time to 10, run the simulation, and examine the plot of (t) in the
Scope. This completes the simulation.
10.7 Subsystems
One potential disadvantage of a graphical interface such as Simulink is that, to
simulate a complex system, the diagram can become rather large and therefore
somewhat cumbersome. Simulink, however, provides for the creation of subsystem
blocks, which play a role analogous to that of subprograms in a programming
language. A subsystem block is actually a Simulink program represented by a
single block. A subsystem block, once created, can be used in other Simulink
programs. We also introduce some other blocks in this section.
To illustrate subsystem blocks, we will use a simple hydraulic system whose
model is based on the conservation of mass principle familiar to engineers.
Because the governing equations are similar to other engineering applications,
such as electric circuits and devices, the lessons learned from this example will
enable you to use Simulink for other applications.
A Hydraulic System
The working uid in a hydraulic system is an incompressible uid such as water
or a silicon-based oil. (Pneumatic systems operate with compressible uids,
such as air.) Consider a hydraulic system composed of a tank of liquid of mass
density (Figure 10.7–1). The tank shown in cross section in the gure is cylindrical with a bottom area A. A ow source dumps liquid into the tank at the mass
ow rate qmi(t). The total mass in the tank is m Ah, and from conservation of
mass we have
dh
dm
= A
= qmi - qmo
(10.7–1)
dt
dt
since and A are constants.
qmi
pa
h
A
R
pa
qmo
Figure 10.7–1 A hydraulic system with a ow source.
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If the outlet is a pipe that discharges to atmospheric pressure pa and provides
a resistance to ow that is proportional to the pressure dif ference across its ends,
then the outlet ow rate is
gh
1
[(gh + pa) - pa] =
R
R
qmo =
where R is called the uid resistance. Substituting this expression into Equation (10.7–1) gives the model
A
g
dh
= qmi(t) h
dt
R
(10.7–2)
The transfer function is
H(s)
1
=
Qmi(s)
As + g>R
On the other hand, the outlet may be a valve or other restriction that provides
nonlinear resistance to the ow . In such cases, a common model is the signedsquare-root (SSR) relation
qmo =
1
SSR(p)
R
where qmo is the outlet mass ow rate, R is the resistance, p is the pressure
difference across the resistance, and
SSR(p) = e
1p
- 1ⱍpⱍ
if p Ú 0
if p 6 0
Note that we may express the SSR(u) function in MATLAB as follows:
sgn(u)*sqrt(abs(u)).
Consider the slightly different system shown in Figure 10.7–2, which has a
ow source q and two pumps that supply liquid at the pressures pl and pr. Suppose
q
pa
A
pl
Rl
h
p
Rr
pr
Figure 10.7–2 A hydraulic system with a ow source
and two pumps.
pa
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Subsystems
445
the resistances are nonlinear and obey the signed-square-root relation. Then the
model of the system is
A
1
dh
1
= q + SSR( pl - p) SSR( p - pr)
dt
Rl
Rr
where A is the bottom area and p = gh. The pressures pl and pr are the gauge
pressures at the left- and right-hand sides. Gauge pressure is the difference between
the absolute pressure and atmospheric pressure. Note that the atmospheric pressure pa cancels out of the model because of the use of gauge pressure.
We will use this application to introduce the following Simulink elements:
■
■
Subsystem blocks
Input and output ports
You can create a subsystem block in one of two ways: by dragging the
Subsystem block from the library to the Model window or by rst creating a
Simulink model and then “encapsulating” it within a bounding box. We will
illustrate the latter method.
We will create a subsystem block for the liquid-level system shown in
Figure 10.7–2. First construct the Simulink model shown in Figure 10.7–3. The
oval blocks are Input and Output Ports (In 1 and Out 1), which are available in
the Ports and Subsystems library. Note that you can use MATLAB variables and
expressions when entering the gains in each of the four Gain blocks.
Before running the program, we will assign values to these variables in the
MATLAB Command window. Enter the gains for the four Gain blocks using the
expressions shown in the block. You may also use a variable as the Initial condition of the Integrator block. Name this variable h0.
The SSR blocks are examples of the Fcn block, which is in the User-De ned
Functions library. Double-click on the block and enter the MATLAB expression
+
1
–
Left Pressure
f(u)
SSR
1/R_I
3
Left Resistance
+
–
++
f(u)
SSR1
1
s
rho*g
Integrator
rho*g
1/(rho*A)
1/(rho*A)
+–
2
Right Pressure
2
Liquid
Height
Mass Flow Input
1/R_r
Right Resistance
Figure 10.7–3 Simulink model of the system shown in Figure 10.7–2.
1
Bottom
Pressure
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1
Left Pressure
Left Pressure
Bottom Pressure
2
Right Pressure
Right Pressure
1
Bottom
Pressure
Liquid Height
Mass Flow Input
3
Mass Flow Input
Subsystem
2
Liquid
Height
Figure 10.7–4 The Subsystem block.
sgn(u)*sqrt(abs(u)). Note that the Fcn block requires you to use the
variable u. The output of the Fcn block must be a scalar, as is the case here, and
you cannot perform matrix operations in the Fcn block, but these are not needed
here. (An alternative to the Fcn block is the MATLAB Fcn block to be discussed
in Section 10.9.) Save the model and give it a name, such as Tank.
Now create a “bounding box” surrounding the diagram. Do this by placing
the mouse cursor in the upper left, holding the mouse button down, and dragging
the expanding box to the lower right to enclose the entire diagram. Then choose
Create Subsystem from the Edit menu. Simulink will then replace the diagram
with a single block having as many input and output ports as required and will
assign default names. You can resize the block to make the labels readable. You
may view or edit the subsystem by double-clicking on it. The result is shown in
Figure 10.7–4.
Connecting Subsystem Blocks
We now create a simulation of the system shown in Figure 10.7–5, where the mass
in ow rate q is a step function. To do this, create the Simulink model shown in
Figure 10.7–6. The square blocks are Constant blocks from the Sources library.
These give constant inputs (which are not the same as step function inputs).
The larger rectangular blocks are two subsystem blocks of the type just created. To insert them into the model, open the Tank subsystem model, select Copy
from the Edit menu, then paste it twice into the new model window. Connect the
q1
h1
A1
A2
R1
Figure 10.7–5 A hydraulic system with two tanks.
h2
R2
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10.7
0
Subsystems
Left Pressure
Bottom Pressure
No Left
Input
Right Pressure
Liquid Height
Mass Flow Input
Mass Inflow
Tank 1
Left Pressure
Bottom Pressure
0
Atmosphere
Scope
Right Pressure
0
Liquid Height
Mass Flow Input
Input Flow 2
Tank 2
Figure 10.7–6 Simulink model of the system shown in Figure 10.7–5.
input and output ports and edit the labels as shown. Then double-click on the
Tank 1 subsystem block, set the left-side gain 1/R_l equal to 0, the right-side
gain 1/R_r equal to 1/R_1, and the gain 1/rho*A equal to 1/rho*A_1. Set
the Initial condition of the integrator to h10. Note that setting the gain 1/R_l
equal to 0 is equivalent to R_l q , which indicates no inlet on the left-hand side.
Then double-click on the Tank 2 subsystem block, set the left-side gain
1/R_l equal to 1/R_1, the right-side gain 1/R_r equal to 1/R_2, and the
gain 1/rho*A equal to 1/rho*A_2. Set the Initial condition of the integrator
to h20. For the Step block, set the Step time to 0, the Initial value to 0, the Final
value to the variable q_1, and the Sample time to 0. Save the model using a
name other than Tank.
Before you run the model, in the Command window assign numerical values
to the variables. As an example, you may type the following values for water, in
U.S. Customary units, in the Command window.
>>A_1 = 2;A_2 = 5;rho = 1.94;g = 32.2;
>>R_1 = 20;R_2 = 50;q_1 = 0.3;h10 = 1;h20 = 10;
After selecting a simulation Stop time, you may run the simulation. The
Scope will display the plots of the heights h1 and h2 versus time.
Figures 10.7–7, 10.7–8, and 10.7–9 illustrate some electrical and mechanical
systems that are likely candidates for application of subsystem blocks. In Figure 10.7–7, the basic element for the subsystem block is an RC circuit. In
Figure 10.7–8, the basic element for the subsystem block is a mass connected to
two elastic elements.
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448
R1
R2
Rn1
Rn
+
C1
v1
C2
Cn1
Cn
vn
–
Figure 10.7–7 A network of RC loops.
k1
k2
m1
k3
kn1
m2
mn
Figure 10.7–8 A vibrating system.
Load
Torque
Control
Voltage
K1
1
Ls R
KT
1
Is c
Speed
K2
Figure 10.7–9 An armature-controlled dc motor.
Figure 10.7–9 is the block diagram of an armature-controlled dc motor,
which may be converted to a subsystem block. The inputs for the block would be
the voltage from a controller and a load torque, and the output would be the
motor speed. Such a block would be useful in simulating systems containing several motors, such as a robot arm.
10.8 Dead Time in Models
TRANSPORT
DELAY
Dead time, also called transport delay, is a time delay between an action and its
effect. It occurs, for example, when a uid ows through a conduit. If the uid
velocity ␷ is constant and the conduit length is L, it takes a time T = L>␷ for the
uid to move from one end to the other . The time T is the dead time.
Let 1(t) denote the incoming uid temperature and 2(t) the temperature of
the uid leaving the conduit. If no heat ener gy is lost, then 2(t) = 1(t - T).
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10.8
Dead Time in Models
From the shifting property of the Laplace transform,
2(s) = e-Ts 1(s)
So the transfer function for a dead-time process is e-Ts.
Dead time may be described as a “pure” time delay, in which no response at
all occurs for a time T, as opposed to the time lag associated with the time constant of a response, for which 2(t) = (1 - e-t>) 1(t).
Some systems have an unavoidable time delay in the interaction between components. The delay often results from the physical separation of the components
and typically occurs as a delay between a change in the actuator signal and its effect
on the system being controlled, or as a delay in the measurement of the output.
Another, perhaps unexpected, source of dead time is the computation time required for a digital control computer to calculate the control algorithm. This can be
a signi cant dead time in systems using inexpensive and slower microprocessors.
The presence of dead time means the system does not have a characteristic
equation of nite order . In fact, there are an in nite number of characteristic
roots for a system with dead time. This can be seen by noting that the term e-Ts
can be expanded in an in nite series as
1
1
e-Ts = Ts =
e
1 + Ts + T 2s2>2 + Á
The fact that there are an in nite number of characteristic roots means that the
analysis of dead-time processes is dif cult, and often simulation is the only
practical way to study such processes.
Systems having dead-time elements are easily simulated in Simulink. The
block implementing the dead-time transfer function e-Ts is called the Transport
Delay block.
Consider the model of the height h of liquid in a tank, such as that shown in
Figure 10.7–1, whose input is a mass ow rate qi. Suppose that it takes a time T
for the change in input ow to reach the tank following a change in the valve
opening. Thus, T is a dead time. For speci c parameter values, the transfer function has the form
H(s)
2
= e-Ts
Qi(s)
5s + 1
Figure 10.8–1 shows a Simulink model for this system. After placing the Transport
Delay block, set the delay to 1.25. Set the Step time to 0 in the Step Function
block. We will now discuss the other blocks in the model.
Specifying Initial Conditions with Transfer Functions
The “Transfer Fcn (with initial outputs)” block, so-called to distinguish it from
the Transfer Fcn block, enables us to set the initial value of the block output. In
our model, this corresponds to the initial liquid height in the tank. This feature
thus provides a useful improvement over traditional transfer-function analysis, in
which initial conditions are assumed to be zero.
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450
4
Gain
+–
Unit-Step
Command
2
5s1
+
+
5/4
s
Rate Saturation Transport Transfer Fcn
Limiter
(with initial
Delay
outputs)
Height
Transfer Fcn
Figure 10.8–1 Simulink model of a hydraulic system with dead time.
The Transfer Fcn (with initial outputs) block is equivalent to adding the free
response to the block output, with all the block’s state variables set to zero except
for the output variable. The block also lets you assign an initial value to the block
input, but we will not use this feature and so will leave the Initial input set to 0 in
the Block Parameters window. Set the Initial output to 0.2 to simulate an initial
liquid height of 0.2.
The Saturation and Rate Limiter Blocks
Suppose that the minimum and maximum ow rates available from the input ow
valve are 0 and 2. These limits can be simulated with the Saturation block, discussed in Section 10.4. After placing the block as shown in Figure 10.8–1, doubleclick on it and type 2 in its Upper limit window and 0 in the Lower limit window.
In addition to being limited by saturation, some actuators have limits on how
fast they can react. This limitation might be due to deliberate restrictions placed on
the unit by its manufacturer to avoid damage to the unit. An example is a ow control valve whose rate of opening and closing is controlled by a rate limiter.
Simulink has such a block, and it can be used in series with the Saturation block to
model the valve behavior. Place the Rate Limiter block as shown in Figure 10.8–1.
Set the Rising slew rate to 1 and the Falling slew rate to 1.
A Control System
PI CONTROLLER
The Simulink model shown in Figure 10.8–1 is for a speci c type of control
system called a PI controller, whose response f (t) to the error signal e(t) is the
sum of a term proportional to the error signal and a term proportional to the integral of the error signal. That is,
t
f (t) = KP e(t) + KI
e(t) dt
L0
where KP and KI are called the proportional and integral gains. Here the error
signal e(t) is the difference between the unit-step command representing the
desired height and the actual height. In transform notation this expression becomes
KI
KI
E(s) = a Kp +
b E(s)
F(s) = KP E(s) +
s
s
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Simulation of a Nonlinear Vehicle Suspension Model
In Figure 10.8–1, we used the values KP 4 and KI 5兾4. These values are
computed using the methods of control theory. (For a discussion of control systems, see, for example, [Palm, 2010].) The simulation is now ready to be run. Set
the Stop time to 30 and observe the behavior of the liquid height h(t) in the
Scope. Does it reach the desired height of 1?
10.9 Simulation of a Nonlinear Vehicle
Suspension Model
Linear or linearized models are useful for predicting the behavior of dynamic
systems because powerful analytical techniques are available for such models,
especially when the inputs are relatively simple functions such as the impulse,
step, ramp, and sine. Often in the design of an engineering system, however, we
must eventually deal with nonlinearities in the system and with more complicated inputs such as trapezoidal functions, and this must often be done with
simulation.
In this section we introduce four additional Simulink elements that enable us
to model a wide range of nonlinearities and input functions:
■ Derivative block
■ Signal Builder block
■ Look-Up Table block
■ MATLAB Fcn block
As our example, we will use the single-mass suspension model shown in
Figure 10.9–1, where the spring and damper forces fs and fd have the nonlinear
models shown in Figures 10.9–2 and 10.9–3. The damper model is unsymmetric
and represents a damper whose force during rebound is higher than during
jounce (in order to minimize the force transmitted to the passenger compartment
when the vehicle strikes a bump). The bump is represented by the trapezoidal
function y(t) shown in Figure 10.9–4. This function corresponds approximately to
a vehicle traveling at 30 mi/h over a road surface elevation 0.2 m high and 48 m
long.
Body
m
m
x
c
k
Road
Suspension
fs
fd
y
Datum level
(a)
(b)
Figure 10.9–1 Single-mass model of a vehicle suspension.
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Spring force (N)
4500
0.5
0.3
500
500
0.3
Deflection
y x (m)
0.5
4500
Figure 10.9–2 Nonlinear spring function.
f = 200v 0.6
0
Damper force f (N)
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0
v = dy/dt – dx /dt (m/s)
Figure 10.9–3 Nonlinear damping function.
y(t) (m)
0.2
0.1
3.0
Figure 10.9–4 Road surface pro le.
3.1
t (s)
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10.9
d
dt
Page 453
453
fd
y
fs
1
m
1
s
1
s
x
x
Figure 10.9–5 Simulation diagram of a vehicle suspension model.
The system model from Newton’s law is
.
.
..
mx = fs (y - x) + fd (y - x )
where m 400 kg, fs (yx) is the nonlinear spring function shown in Fig.
.
ure 10.9–2, and fd (y - x ) is the nonlinear damper function shown in Figure 10.9–3. The corresponding simulation diagram is shown in Figure 10.9–5.
The Derivative and Signal Builder Blocks
.
The simulation diagram shows that we need to compute y . Because Simulink
uses numerical and not analytical methods, it computes derivatives only approximately, using the Derivative block. We must keep this in mind when using
rapidly changing or discontinuous inputs. The Derivative block has no settings,
so merely place it in the Simulink diagram as shown in Figure 10.9–6.
du/dt
+–
Derivative
MATLAB
Function
Signal 1
Signal Builder
+
+
+
–
Look-Up
Table
1
s
1/400
1/m
1
s
Integrator Integrator 1
Scope
simout
To
Workspace
Clock
–
+
Figure 10.9–6 Simulink model of a vehicle suspension system.
Scope 1
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Next, place the Signal-Builder block, then double-click on it. A plot window
appears that enables you to place points to de ne the input function. Follow the
directions in the window to create the function shown in Figure 10.9–4.
The Look-Up Table Block
The spring function fs is created with the Look-Up Table block. After placing it
as shown, double-click on it and enter [-0.5, -0.1, 0, 0.1, 0.5] for
the Vector of input values and [-4500, -500, 0, 500, 4500] for the
Vector of output values. Use the default settings for the remaining parameters.
Place the two integrators as shown, and make sure the initial values are set
to 0. Then place the Gain block and set its gain to 1兾400. The To Workspace
block and the Clock will enable you to plot x(t) and y(t)x(t) versus t in the
MATLAB Command window.
The MATLAB Fcn Block
In Section 10.7 we used the Fcn block to implement the signed-square-root function. We cannot use this block for the damper function shown in Figure 10.9–3
because we must write a user-de ned function to describe it. This function is as
follows.
function f = damper(v)
if v <= 0
f = -800*(abs(v)).^(0.6);
else
f = 200*v.^(0.6);
end
Create and save this function le. After placing the MATLAB Fcn block,
double-click on it and enter its name damper. Make sure Output dimensions is
set to 1 and the Output signal type is set to auto.
The Fcn, MATLAB Fcn, Math Function, and S-Function blocks can be used
to implement functions, but each has its advantages and limitations. The Fcn
block can contain an expression, but its output must be a scalar, and it cannot call
a function le. The MATLAB Fcn block is slower than the Fcn block, but its output can be an array, and it can call a function le.
The Math Function block can produce an array output, but it is limited to
a single MATLAB function and cannot use an expression or call a le. The
S-Function block provides more advanced features, such as the ability to use
C language code.
The Simulink model when completed should look like Figure 10.9–6. You
can plot the response x(t) in the Command window as follows:
>>x = simout(:,1);
>>t = simout(:,3);
>>plot(t,x),grid,xlabel(‘t (s)’),ylabel(‘x (m)’)
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10.10
Summary
0.3
0.25
0.2
0.15
x (m)
0.1
0.05
0
–0.05
–0.1
–0.15
–0.2
0
0.5
1
1.5
2
2.5
t (s)
3
3.5
4
4.5
5
Figure 10.9–7 Output of the Simulink model shown in Figure 10.9–6.
The result is shown in Figure 10.9–7. The maximum overshoot is seen to be
0.26 0.2 0.06 m, but the maximum undershoot is seen to be much greater,
0.168 m.
10.10 Summary
The Simulink model window contains menu items we have not discussed. However, the ones we have discussed are the most important ones for getting started.
We have introduced just a few of the blocks available within Simulink. Some of
the blocks not discussed deal with discrete-time systems (ones modeled with difference, rather than differential, equations), digital logic systems, and other types
of mathematical operations. In addition, some blocks have additional properties
that we have not mentioned. However, the examples given here will help you get
started in exploring the other features of Simulink. Consult the online help for information about these items.
Key Terms with Page References
Block diagram, 420
Dead time, 448
Dead zone, 437
Derivative block, 453
Fcn block, 454
Gain block, 420
Integrator block, 420
Library Browser, 421
Look-Up Table block, 454
PI controller, 450
Piecewise-linear models, 430
Rate Limiter block, 450
455
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Relay block, 433
Saturation block, 450
Signal Builder block, 453
Simulation diagrams, 420
State-variable models, 441
Subsystems, 443
Summer, 421
Transfer-function models, 437
Transport delay, 448
Problems
Section 10.1
1.
Draw a simulation diagram for the following equation.
.
y = 5f (t) - 7y
2.
Draw a simulation diagram for the following equation.
..
.
5y = 3y + 7y = f (t)
3.
Draw a simulation diagram for the following equation.
.
3y + 5 sin y = f(t)
Section 10.2
4.
Create a Simulink model to plot the solution of the following equation for
0 … t … 6.
.
..
10y = 7 sin 4t + 5 cos 3t y(0) = 3 y (0) = 2
5.
A projectile is launched with a velocity of 100 m/s at an angle of 30
above the horizontal. Create a Simulink model to solve the projectile’s
equations of motion where x and y are the horizontal and vertical displacements of the projectile.
.
..
x= 0
x(0) = 0 x (0) = 100 cos 30°
..
.
y = - g y(0) = 0 y (0) = 100 sin 30°
Use the model to plot the projectile’s trajectory y versus x for
0 … t … 10 s.
6.
The following equation has no analytical solution even though it is linear.
.
x + x = tan t x(0) = 0
The approximate solution, which is less accurate for large values of t, is
x(t) =
1 3
t - t2 + 3t - 3 + 3e - t
3
Create a Simulink model to solve this problem, and compare its solution
with the approximate solution over the range 0 … t … 1.
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Problems
7.
Construct a Simulink model to plot the solution of the following equation
for 0 … t … 10
#
15x + 5x = 4us(t) - 4us(t - 2) x(0) = 5
where us(t) is a unit-step function (in the Block Parameters window of the
Step block, set the Step time to 0, the Initial value to 0, and the Final
value to 1).
8.
A tank having vertical sides and a bottom area of 100 ft2 is used to store
water. To ll the tank, water is pumped into the top at the rate given in the
following table. Use Simulink to solve for and plot the water height h(t)
for 0 t 10 min.
Time (min)
3
Flow Rate (ft /min)
0
1
2
3
4
5
6
7
8
9
10
0
80
130
150
150
160
165
170
160 140
120
Section 10.3
9.
Construct a Simulink model to plot the solution of the following equations for 0 t 2
.
x1 = - 6x1 + 4x2
.
x2 = 5x1 - 7x2 + f(t)
where f(t) 3t. Use the Ramp block in the Sources library.
10.
Construct a Simulink model to plot the solution of the following equations for 0 t 3
.
x1 = - 6x1 + 4x2 + f1(t)
.
x2 = 5x1 - 7x2 + f2(t)
where f1(t) is a step function of height 3 starting at t 0 and f2(t) is a step
function of height 3 starting at t 1.
Section 10.4
11.
Use the Saturation block to create a Simulink model to plot the solution
of the following equation for 0 t 6.
.
y(0) = 3
3y + y = f(t)
where
8
f (t) =
-8
L 10 sin 3t
if 10 sin 3t > 8
if 10 sin 3t < - 8
otherwise
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12.
Construct a Simulink model of the following problem.
.
5x + sin x = f(t)
x(0) = 0
The forcing function is
-5
f(t) = g(t)
L5
if g(t) … -5
if - 5 … g(t) … 5
if g(t) Ú 5
where g(t) 10 sin 4t.
13.
If a mass-spring system has Coulomb friction on the surface rather than
viscous friction, its equation of motion is
.
- ky + f (t) - mg if y Ú 0
..
my = e
.
- ky + f (t) + mg if y < 0
where is the coef cient of friction. Develop a Simulink model for the
case where m 1kg, k 5 N/m, 0.4, and g 9.8 m/s2. Run the
simulation for two cases: (a) the applied force f(t) is a step function with
a magnitude of 10 N and (b) the applied force is sinusoidal: f (t) 10 sin 2.5t. Either the Sign block in the Math Operations library or the
Coulomb and Viscous Friction block in the Discontinuities library can be
used, but since there is no viscous friction in this problem, the Sign block
is easier to use.
14.
A certain mass, m 2 kg, moves on a surface inclined at an angle
= 30° above the horizontal. Its initial velocity is ␷ (0) 3 m/s up the
incline. An external force of f1 = 5 N acts on it parallel to and up the incline. The coef cient of Coulomb friction is = 0.5. Use the Sign block
and create a Simulink model to solve for the velocity of the mass until the
mass comes to rest. Use the model to determine the time at which the
mass comes to rest.
15.
a. Develop a Simulink model of a thermostatic control system in which
the temperature model is
RC
dT
+ T = Rq + Ta (t)
dt
where T is the room air temperature in F, Ta is the ambient (outside)
air temperature in F, time t is measured in hours, q is the input from
the heating system in lb # ft/hr, R is the thermal resistance, and C is the
thermal capacitance. The thermostat switches q on at the value qmax
whenever the temperature drops below 69 F and switches q to q 0
whenever the temperature is above 71 F. The value of qmax indicates
the heat output of the heating system.
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Problems
Run the simulation for the case where T(0) 70 F, and Ta(t) 50 10 sin (t兾12). Use the values R 5 105 F # hr/lb # ft and C 4 104 lb # ft/°F. Plot the temperatures T and Ta versus t on the same
graph, for 0 t 24 hr. Do this for two cases: qmax 4 105 and
qmax 8 105 lb # ft/ hr. Investigate the effectiveness of each case.
b. The integral of q over time is the energy used. Plot 1 q dt versus t
and determine how much energy is used in 24 hr for the case where
qmax 8 105.
16.
Refer to Problem 15. Use the simulation with qmax 8 105 to compare
the energy consumption and the thermostat cycling frequency for the two
temperature bands (69 , 71 ) and (68 , 72 ).
17.
Consider the liquid-level system shown in Figure 10.7–1. The governing
equation based on conservation of mass is Equation (10.7–2). Suppose
that the height h is controlled by using a relay to switch the input ow
rate between the values 0 and 50 kg/s. The ow rate is switched on when
the height is less than 4.5 m and is switched off when the height reaches
5.5 m. Create a Simulink model for this application using the values
A 2 m2, R 400 m1 # s1, 1000 kg/m3, and h(0) 1 m. Obtain
a plot of h(t).
Section 10.5
18.
Use the Transfer Function block to construct a Simulink model to plot the
solution of the following equation for 0 t 4.
.
..
.
2x + 12x + 10x = 5us(t) - 5us(t - 2) x(0) = x (0) = 0
19.
Use Transfer Function blocks to construct a Simulink model to plot the
solution of the following equations for 0 t 2.
..
.
.
3x + 15x + 18x = f(t) x(0) = x (0) = 0
.
..
.
2y + 16y + 50y = x(t) y(0) = y (0) = 0
where f(t) 75us(t).
20.
Use Transfer Function blocks to construct a Simulink model to plot the
solution of the following equations for 0 t 2
.
..
.
3x + 15x + 18x = f(t) x(0) = x (0) = 0
.
..
.
2y + 16y + 50y = x(t) y(0) = y (0) = 0
where f(t) 50us (t). At the output of the rst block there is a dead zone
for 1 x 1. This limits the input to the second block.
21.
Use Transfer Function blocks to construct a Simulink model to plot the
solution of the following equations for 0 t 2
.
..
.
3x + 15x + 18x = f(t) x(0) = x (0) = 0
.
..
.
2y + 16y + 50y = x(t) y(0) = y (0) = 0
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CHAPTER 10 Simulink
where f(t) 50us (t). At the output of the rst block there is a saturation
that limits x to be x 1. This limits the input to the second block.
Section 10.6
22.
Construct a Simulink model to plot the solution of the following equation
for 0 t 4.
.
..
.
2x + 12x + 10x 2 = 8 sin 0.8t x(0) = x (0) = 0
23.
Create a Simulink model to plot the solution of the following equation for
0 t 3.
#
x + 10x2 = 5 sin 3t x(0) = 1
24.
Construct a Simulink model of the following problem.
.
10x + sin x = f(t) x(0) = 0
The forcing function is f (t) sin 2t. The system has the dead-zone nonlinearity shown in Figure 10.5–1.
25.
The following model describes a mass supported by a nonlinear, hardening spring. The units are SI. Use g 9.81 m/s2.
.
..
5y = 5g - (900y + 1700y3 ) y(0) = 0.5 y (0) = 0
Create a Simulink model to plot the solution for 0 t 2.
26.
Consider the system for lifting a mast shown in Figure P26. The 70-ft-long
mast weighs 500 lb. The winch applies a force f 380 lb to the cable.
The mast is supported initially at an angle of 30 , and the cable at A is initially horizontal. The equation of motion of the mast is
..
626 000
25 400 = -17 500 cos +
sin(1.33 + )
Q
where
Q = 12020 + 1650 cos(1.33 + )
Create and run a Simulink model to solve for and plot (t) for (t) 兾2 rad.
d
D
A
380 lb
H 20
O
W 5
Figure P26
30°
L 40
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Problems
27.
The equation describing the water height h in a spherical tank with a drain
at the bottom is
dh
= - Cd A 12gh
(2rh - h2)
dt
Suppose that the tank’s radius is r 3 m and the circular drain hole of
area A has a radius of 2 cm. Assume that Cd 0.5 and that the initial
water height is h(0) 5 m. Use g 9.81 m/s2. Use Simulink to solve the
nonlinear equation, and plot the water height as a function of time until
h(t) 0.
28.
A cone-shaped paper drinking cup (like the kind used at water fountains)
has a radius R and a height H. If the water height in the cup is h, the water
volume is given by
R 2
1
V = a b h3
3
H
Suppose that the cup’s dimensions are R1.5 in. and H 4 in.
a. If the ow rate from the fountain into the cup is 2 in. 3/sec, use
Simulink to determine how long will it take to ll the cup to the brim.
b. If the ow rate from the fountain into the cup is given by 2(1 e2t )
in.3/sec, use Simulink to determine how long will it take to ll the cup
to the brim.
Section 10.7
29.
Refer to Figure 10.7–2. Assume that the resistances obey the linear relation, so that the mass ow ql through the left-hand resistance is ql (pl – p)兾Rl, with a similar linear relation for the right-hand resistance.
a. Create a Simulink subsystem block for this element.
b. Use the subsystem block to create a Simulink model of the system
shown in Figure 10.7–5. Assume that the mass in ow rate is a step
function.
c. Use the Simulink model to obtain plots of h1(t) and h2(t) for the following parameter values: A1 = 2 m2, A2 = 5 m2, R1 = 400 m1 # s1
R2 = 600 m1 # s1, = 1000 kg/m3, qmi = 50 kg/s, h1(0) = 1.5 m,
and h2(0) = 0.5 m.
30.
a. Use the subsystem block developed in Section 10.7 to construct a
Simulink model of the system shown in Figure P30. The mass in ow
rate is a step function.
b. Use the Simulink model to obtain plots of h1(t) and h2(t) for the
following parameter values: A1 3 ft2, A2 5 ft2, R1 30 ft1 # sec1,
R2 40 ft1 # sec1, = 1.94 slug/ft3, qmi 0.5 slug/sec, h1(0) 2 ft,
and h2(0) 5 ft.
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qmi
h1
R1
A1
h2
R2
A2
Figure P30
31.
Consider Figure 10.7–7 for the case where there are three RC loops with
the values R1 = R3 = 104 , R2 = 5 * 104 , C1 = C3 = 10-6 F, and
C2 = 4 * 10-6 F.
a. Develop a subsystem block for one RC loop.
b. Use the subsystem block to construct a Simulink model of the
entire system of three loops. Plot ␷3(t) over 0 … t … 3 for
␷1(t) = 12 sin 10t V.
32.
Consider Figure 10.7–8 for the case where there are three masses. Use
the values m1 m3 10 kg, m2 30 kg, k1 k4 104 N/m, and
k2 k3 2 104 N/m.
a. Develop a subsystem block for one mass.
b. Use the subsystem block to construct a Simulink model of the entire
system of three masses. Plot the displacements of the masses over
0 … t … 2 s for if the initial displacement of m1 is 0.1 m.
Section 10.8
33.
Refer to Figure P30. Suppose there is a dead time of 10 sec between the
out ow of the top tank and the lower tank. Use the subsystem block
developed in Section 10.7 to create a Simulink model of this system.
Using the parameters given in Problem 30, plot the heights h1 and h2
versus time.
Section 10.9
34.
Redo the Simulink suspension model developed in Section 10.9, using the
spring relation and input function shown in Figure P34 and the following
damper relation.
fd (␷) = e
-500ⱍ␷ⱍ1.2
50␷1.2
␷ … 0
␷ 7 0
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Problems
463
Spring force (N)
3000
y(t)
(m)
0.3
1300
0.25
0.15
500
500 0.15
0.25
Deflection
y x (m)
0
0
0.15
4.0 4.15 t (s)
(b)
1300
3000
(a)
Figure P34
Use the simulation to plot the response. Evaluate the overshoot and
undershoot.
35.
Consider the system shown in Figure P35. The equations of motion are
$
#
#
m1x 1 + (c1 + c2)x1 + (k1 + k2)x1 - c2 x2 - k2 x2 = 0
$
#
#
m2 x 2 + c2 x2 + k2 x2 - c2 x1 - k2 x1 = f(t)
Suppose that m1 = m2 = 1, c1 = 3, c2 = 1, k1 = 1, and k2 = 4.
a. Develop a Simulink model of this system. In doing this, consider
whether to use a state-variable representation or a transfer-function
representation of the model.
b. Use the Simulink model to plot the response x1(t) for the following
input. The initial conditions are zero.
t
f(t) = 2 - t
L0
0 … t … 1
1 6 t 6 2
t Ú 2
c2
c1
f (t)
m2
Figure P35
k2
m1
k1
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© Royalty-Free/CORBIS
Engineering in the
21st Century. . .
Developing Alternative
Sources of Energy
I
t now appears that the United States and much of the rest of the world have
recognized the need to reduce their dependence on nonrenewable energy
sources such as natural gas, oil, coal, and perhaps even uranium. Supplies of
these fuels will eventually be exhausted, they have harmful environmental effects,
and when imported, they cause huge trade imbalances that hurt the economy.
One of the major engineering challenges of the 21st century will be to develop
renewable energy sources.
Renewable energy sources include solar energy (both solar-thermal and
solar-electric), geothermal power, tidal and wave power, wind power, as well as
crops that can be converted to alcohol. In solar-thermal applications, energy from
the sun is used to heat a uid, which can be used to heat a building or power an
electrical generator such as a steam turbine. In solar-electric applications, sunlight is directly converted to electricity.
Geothermal power is obtained from ground heat or steam vents. With tidal
power the tidal currents are used to drive a turbine to generate electricity. Wave
power uses the change in water surface level due to waves to drive water through
a turbine or other device. Wind power uses a wind turbine to drive a generator.
The dif culty with most renewable ener gy sources is that they are diffuse, so
the energy must be concentrated somehow, and they are intermittent, which
requires a storage method. At present most renewable energy systems are not
very ef cient, and so the engineering challenge of the future is to improve their
ef ciency . ■
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C H A P T E R
11
MuPAD
OUTLINE
11.1 Introduction to MuPAD
11.2 Symbolic Expressions and Algebra
11.3 Algebraic and Transcendental Equations
11.4 Linear Algebra
11.5 Calculus
11.6 Ordinary Differential Equations
11.7 Laplace Transforms
11.8 Special Functions
11.9 Summary
Problems
MuPAD is a very large program with many capabilities. In this chapter we cover
a subset of those capabilities, emphasizing the ones of greatest use to beginning
students in engineering and science. Speci cally we treat the following:
■
■
■
■
■
■
■
■
MuPAD basics and the Help system.
Symbolic algebra.
Methods for solving algebraic and transcendental equations.
Selected topics in linear algebra.
Symbolic methods for solving ordinary differential equations.
Symbolic calculus.
Laplace transforms.
The special functions of mathematics.
465
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CHAPTER 11 MuPAD
When you have nished this chapter , you should be able to use MATLAB to do
the following:
■
■
■
■
■
■
■
■
Create symbolic expressions and manipulate them algebraically.
Obtain symbolic and numeric solutions to algebraic and transcendental
equations.
Perform symbolic linear algebra operations, including obtaining expressions for determinants, matrix inverses, eigenvectors, and eigenvalues.
Perform symbolic differentiation and integration.
Evaluate limits and series symbolically.
Obtain symbolic solutions to ordinary differential equations.
Obtain and apply Laplace transforms.
Solve ordinary differential equations in terms of special functions or series.
11.1 Introduction to MuPAD
MuPAD is a big package with a lot of capabilities. In this chapter we introduce a
subset of those capabilities, ones that are most useful to engineers and scientists,
and only the basic syntax of those commands. Most commands in MuPAD have
an extended syntax that is documented in the MuPAD Help system. You will
need to refer frequently to this Help system, because space limitations make it
impossible to cover MuPAD in detail in one chapter.
MuPAD comes with the Symbolic Math toolbox. The toolbox itself has its
own syntax, different somewhat from that of MuPAD, and its commands are entered in the Command window or in M- les. The MuPAD interface, however, is a
notebook. Prior to MATLAB Release 2008b (Version 5.1 of the toolbox), the toolbox used a licensed version of Maple as its engine. Now the Maple engine has
been replaced by the MuPAD engine, but the toolbox syntax remains the same.
However, the MuPAD engine behaves somewhat differently from the Maple
engine. One example of the differing behavior is the solution of transcendental
equations. Whereas the Maple engine sometimes correctly gave more than one
solution, the MuPAD engine often gives only one of the possible solutions. Thus
previous users of the toolbox will nd that their old program may behave
differently now.
In this chapter we cover MuPAD exclusively, because of page length limitations and because MuPAD appears to be the future of the Symbolic Math
toolbox.
The MuPAD Welcome Screen
In MuPAD you perform operations in “notebooks.” Thus you can organize your
work by topics. To start MuPAD, rst start MA TLAB; then to open a new notebook, type mupadwelcome at the MATLAB prompt. This opens the screen
shown in Figure 11.1–1. It enables you to get help under the subwindow First
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11.1
Introduction to MuPAD
Figure 11.1–1 The MuPAD welcome screen.
Figure 11.1–2 The Getting Started screen.
Steps. Clicking on Getting Started opens the screen shown in Figure 11.1–2,
which has a variety of Help topics. The left-hand pane has a list of topics that can
be expanded or collapsed by clicking on the or signs. The right-hand pane
contains a shorter list of topics.
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CHAPTER 11 MuPAD
Figure 11.1–3 The Notebook Interface Help screen.
Referring to the welcome screen shown in Figure 11.1–1, when you click on
Notebook Interface, the screen shown in Figure 11.1–3 appears. It provides
access to the Help features speci c to the Notebook Interface.
You can start a new notebook by clicking on New Notebook in the welcome screen, or you can open a previously created notebook by clicking on
Open Recent File or by clicking on Open File to access les not in the default
directory.
You can avoid the welcome screen by typing mupad at the MATLAB
prompt. This opens a new notebook. Once MuPAD opens, you can select a previously created notebook if you wish.
The Menu Bar
INPUT, TEXT, AND
OUTPUT REGIONS
Figure 11.1–4 shows a previously created notebook containing commands, outputs (the results of executing the commands), and comments. Commands and
comments are placed in input and text regions, respectively. Results appear in output regions. The results can be mathematical expressions, numbers, or graphs.
In the default desktop arrangement, the Command bar appears to the right of
the screen. This bar will be discussed in Section 11.2. You can con gure the
desktop to display the Search and Replace bar instead, but the Command bar is
more useful in general.
The Menu bar is at the top of the interface. It contains nine menus, some of
which you will not need to do basic computations. The File, Edit, and Help
menus are similar to those in MATLAB except that the latter accesses the
MuPAD First Steps screen. The View menu lets you con gure the screen by
adding or removing toolbars. The Navigation menu enables you to go back and
forth in the notebook.
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Figure 11.1–4 The MuPAD Notebook Interface.
Figure 11.1–5 The Standard bar.
The Insert menu is very useful because it enables you to insert calculations
and comments (text paragraphs) at any place in the notebook. By clicking on one
of the Evaluate options in the Notebook menu, the results of any newly inserted
calculations will be propagated throughout the notebook.
The Format menu lets you change font sizes and page formatting. The
Window menu lets you switch between notebooks if more than one is open.
You quit a MuPAD notebook by selecting Close on the File menu. Selecting
Exit on the File menu will close MuPAD and return you to MATLAB.
The Standard Toolbar
Below the Menu bar is the Standard toolbar (Figure 11.1–5). This contains shortcut buttons for commonly used operations whose functions are obvious from the
name of the button. The most useful are the Calculation button, which inserts a
new input region below the current cursor position; the Text Paragraph button,
which inserts a text paragraph; and the Evaluate button, which evaluates the calculation in the input region containing the cursor.
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Entering Commands
You enter commands into input regions in the notebook. Figure 11.1–4 shows a
notebook with some comments, commands, and responses displayed. Input and
output regions have a bracket at the left; text regions do not. After you type a
command in a blank input region, press Enter and MuPAD will evaluate the command. The result will be displayed in the next bracket, just below the input region.
An example session that computes cos(3.14) is the following. The command is
cos(3.14) and the output is –0.9999987317. In this chapter we display commands in boldface type, outputs and comments in italic and roman type.
[ > cos(3.14)
[ 0.9999987317
Note that the answer is not exactly 1 because 3.14 is not exactly . For exact
answers use PI, which is the MuPAD symbol for .
[ > cos(PI)
[ 1
RESERVED
SYMBOLS
Some commonly used reserved symbols in MuPAD are PI for , I for the
complex number i = 1 - 1, and E for e, the base of the natural logarithm. You
should not use these as identi ers to represent something other than their reserved
meaning.
You can edit previous inputs by clicking on them and entering or deleting
text, using the keyboard editing keys. Obtain an updated result by pressing
Enter. You can cut and paste from the input region, but output cannot be edited
or copied because it is a graphic. Terminate the command with a colon to suppress the output display. Enter more than one command on a line by separating
them with either a semicolon or a colon.
MuPAD uses the standard arithmetic operators ( * 兾) and follows the
same precedence rules as MATLAB. MuPAD uses function names that are for
the most part identical to those used by MATLAB, for example, sin, tan, exp.
Two notable exceptions are the natural logarithm ln, which is ln in MuPAD, and
the base-10 logarithm, log10, which is log in MuPAD. Answers are exact (not
rounded oating-point numbers) when computing with integers and rational
numbers. For example,
[ > 1 3*7兾2
23
c
2
[ > (3 (7兾2*5))兾(1兾5 2兾9)^2
83025
c
722
[ > sqrt(52)
[ 2113
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Use E or exp(1) for the base of the natural logarithm, although MuPAD displays the results using the lowercase e. Use ln(x) for ln x and log(x) for log x.
[ > 2*ln(E^3兾2)
c 2 ln a
e3
b
2
[ > oat(%)
[ 4.613705639
The oat function returns a oating-point result. You can use % to refer to
the previous result. MuPAD displays results according to the value of DIGITS,
which sets the number of decimal digits to display. The default value is 10.
Warning! It is easy to make mistakes using the oat function with the % symbol, which represents the last computed result, not necessarily the result just
above the oat(%) command. For example, if you go back several steps in the
notebook and recompute a command, then typing oat(%) later on will cause it
to evaluate the results of that command, even though you type oat(%) as the
last command in the notebook.
Approximate computation is used if at least one of the numbers involved is
a oating-point value. For example, changing the 3 in the earlier example to 3.2
gives a oating-point result.
[ > (3.2 (7兾2*5))兾(1兾5 2兾9)^2
[ 116.1149584
Most of the common mathematical functions will return an exact value if
their argument is exact, and a oating-point value if their ar gument is oatingpoint. For example,
[ > 0.8*cos(2)
[ 0.8 cos(2)
[ > 2兾5*cos(2.1)
[ 0.2019384418
Complex number calculations are easily done using I for 1 - 1.
[ > (1兾3 2*I)*(0.1 I兾2)^3
[ 0.1953333333 0.1846666666i
The functions Re(p) and Im(p) return the real and the imaginary part of a
complex number p. The functions conjugate(p), abs(p), and arg(p) compute the
complex conjugate, the absolute value, and the polar angle of the number p. The
function rectform converts a number to rectangular form.
[ > conjugate(1兾(sqrt(3) I))
1
c
13 - i
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[ > rectform(conjugate(1兾(sqrt(3) I)))
c
i
13
+
4
4
[ > abs(1兾(sqrt(3) I))
c
1
2
[ > arg(1兾sqrt(3) I))
c 6
Inserting Text and Calculations
You can insert text for documentation purposes. An example is shown in Figure 11.1–4. To insert text above a calculation, the easiest way is to position the
cursor within the input region to be documented, then click on the Insert
menu, next on Text Paragraph Above. Then type your text. You may also
want to insert a calculation above another one. To do this, click on the Insert
menu, then on Calculation Above. This opens a blank input region.
11.2 Symbolic Expressions and Algebra
ASSIGNMENT
OPERATOR
Use the assignment operator : to assign a symbolic expression to an identi er
named, say, r. The identi er r can then be used as an abbreviation for the expression. For example,
[ > r: y^2 3*y 5*x 7*x^2 2*x^3
[ 2x3 7x2 5x y2 3y
Notice that the expression has been rearranged in the display.
It is good practice to delete the identi ers of variables, expressions, and
functions to avoid con icts if you later use those identi ers to mean something
else. Use the delete (p) function to delete p. It is also necessary to remove any
assumptions placed on a parameter if that parameter will be used in later computations. Examples of this are the assume and unassume functions, to be discussed later.
Expressions Versus Functions
It is important to realize that an expression in MuPAD is not the same thing as a
function. For example, consider the expression
[ > w : x^6:
We cannot evaluate the expression at a particular value of x, say x 3 by typing
w (3) or by typing oat (w(3)) .
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User-de ned functions can be created by de ning a procedure. Consider the
function f (x) x 2 3x 7. We can de ne it by a procedure as follows:
[ > f : x -> x^2 3*x 7:
The arrow symbol is obtained by typing the hyphen - followed by the greater
than sign (>). We can then evaluate the function at given values of x:
[ > f(x)
[ x2 3x 7
[ > f(y)
[ y 2 3y 7
[ > f(ab)
[ 3a 3b (a b) 2 7
[ > f(2.5)
[ 20.75
You can plot expressions or functions in the same way. For example,
[ > y : 5*exp(-0.5*x)*sin(3*x):
[ > plot (y,x0..5)
The result is shown in Figure 11.2–1. You could also type plot(y) without limits,
but MuPAD would select the plot limits so that the plot was centered around
x 0. Typing plot(%) produces the same result.
Figure 11.2–1 Creating a plot.
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Figure 11.2–2 The Command bar.
Manipulating Expressions
Figure 11.2–2 shows the Command bar. The Command bar provides shortcuts to
avoid typing commands into the input region. When you click on an item, that
command is inserted at the current cursor location in the notebook. Table 11.2–1
lists the operations that are available from the upper portion of the Command bar.
We will discuss these commands throughout the chapter, and you can refer to this
table when particular commands are discussed.
The following functions can be used to manipulate expressions by collecting
coef cients of like powers, expanding powers, and factoring expressions, for
example. You can type functions directly in the input region, or you can select
functions from the Command bar, which contains a subset of all the MuPAD
functions. The following choices appear when you choose General Math from
the Command bar:
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The General Math Menu
Expand
Factor
Normalize
Evaluate
Simplify
Combine
Rewrite
Solve
Table 11.2–1 Items on the Command bar
Derivatives
Integrals
Solve Equations
Numerical Evaluation and Rounding
Math Operators
Trig Functions
Reserved Symbols
Matrices and Vectors
Limits
Rewrite Expressions
Simplify
Equality Tests
Factorials
Exponentials and Logs
Greek Letters
2D Plot
Sums
Products
Evaluate with x a
Assignment
Function De nition
Piecewise De nitions
Physical Units
3D Plot
The last ve items on the General Math menu contain submenus. We will
discuss the Solve function in Section 11.3. When you click on any of these items,
MuPAD inserts the corresponding function into the input region at the location
of the cursor. The function argument will be the place marker #, which you
should delete, and then you enter the appropriate expression. In every case you
may type in the function if you do not want to use the menu.
Expand When you select Expand from the menu, MuPAD inserts expand(#)
in the input region. Delete the place marker # and type in your text. The function
expand(p) expands the expression p by using a set of rules appropriate to the
given expression. The following example shows how the expansion is carried out
by powers.
[ > p1 : (x-5)^2(y-3)^2
[ (x 5)2 (y 3)2
[ > expand(p1)
[ x2 10x y2 6y 34
You can reuse expression symbols such as p1 by typing delete(p1) in the input box.
The next example shows how the expansion is carried out using trigonometric identities.
[ > expand(sin(x y))
[ cos(x) sin(y) cos(y) sin(x)
Factor When you select Factor from the menu, MuPAD inserts factor (#) in
the input region. Use the factor(p) function to factor the expression p.
[ > factor(x^2-1)
[ (x 1)(x 1)
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The factorout(p,q) function factors out a given expression q from the expression p.
[ > factorout((a*b a*c)兾(d*cc), a兾c)
ab + c
c
cd + 1
Normalize Clicking on Normalize places the function normal (#) in the input
area that contains the cursor. The function normal(p) nds a common denominator for rational expressions.
[ > p : x兾(1 x) - 3兾(1 - x)
x
3
c
+
x + 1
x - 1
[ > normal(p)
c
x2 + 2x + 3
x2 - 1
Also, normal cancels common factors in the numerator and the denominator:
[ > x^2兾(x y) - y^2兾(x y)
c
y2
x2
x + y
x + y
[ > normal(%)
[ xy
Evaluate The menu item Evaluate has two subchoices: Numerically and
Boolean. Choosing Numerically inserts oat (#) into the input region. We have
already seen the oat function. Choosing Boolean inserts bool (#), which evaluates a Boolean expression.
[ > bool( oat(sqrt(18)) < oat(sqrt(3)*sqrt(6)))
[ TRUE
Simplify
The Simplify menu item has eight subchoices:
The Simplify Menu
General
Logical
Radical
Relational
Exponential
Logarithm
Sine
Cosine
Choosing General inserts the function Simplify(#). Note the capitalized
rst letter . You can also type the related function simplify(p). The simplify and
Simplify functions are similar; Simplify is more powerful, but may be slower.
The general form of these functions is simplify(p, <target>), where <target>
restricts the simpli cation rules to the tar get function(s). For example, choosing
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the submenu item Logical inserts the function simplify(#, logic). Other choices
for <target> are sqrt, relation, exp, ln, sin, and cos, which are obtained by
choosing the menu items Radical, Relational, Exponential, Logarithm, Sine,
and Cosine, respectively. For example,
[ > simplify(sqrt(4 2*sqrt(3)), sqrt)
[ 13 + 1
Sometimes the target function is not needed. For example,
[ > simplify(cos(x)*sin(y) cos(y)*sin(x))
[ sin(x y)
Combine The Combine menu item has eight subchoices:
The Combine Menu
General
Arc Tangent
Exponential
Logarithm
Power
Sine兾Cosine
Sine兾Cosine Hyp
Square Root
Choosing General inserts the function combine(#) into the notebook. The
combine(p) function combines terms in the expression p of the same algebraic
structure. The general form of the combine function is combine(p <target>).
The <target> option combines several calls to the target functions(s) in the expression p to a single call. Other target functions are available, but are not on the
menu; these are log and gamma.
For example, choosing the submenu item Sine兾Cosine inserts the function
combine(#, sincos). Other choices for <target> are arctan, exp, ln, power,
sincos, sinhcosh, and sqrt, which are obtained by choosing the corresponding
menu items.
When the second argument is not speci ed, combine groups powers of the
same base:
[ > combine(x*y*x^y)
[ xy + 1y
The function combine combines powers with the same exponent in certain
cases, but not always:
[ > combine(sqrt(2)*sqrt(3)*y^5*x^5)
[ 16x5y5
With the target function sincos, products of sines and cosines are expressed
as sums of sines and cosines.
[ > combine(cos(a)*sin(b) sin(a)^2, sincos)
sin(a + b)
cos(2a)
sin(a - b)
1
c
+
2
2
2
2
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Some combinations may require certain assumptions to be made. For example,
[ > assume(a > 0): assume(b > 0):
[ > combine(ln(a) ln(b), ln)
[ ln(ab)
[ > unassume(a): unassume(b):
Rewrite
The Rewrite menu under General Math has eight subchoices:
The Rewrite Menu
Differential
Exponential
Factorial
Gamma
Heaviside
Logarithm
Sign
Sine兾Cosine
Choosing any one of these items inserts the function rewrite(#, <target>)
into the notebook. The rewrite(p, <target>) function transforms an expression p
to a mathematically equivalent form, trying to express p in terms of the speci ed
target function. For example, choosing the submenu item Sine兾Cosine inserts
the function rewrite(#, sincos).
[ > rewrite(cot(x), sincos)
cos(x)
c
sin(x)
[ > rewrite(arcsinh(x), ln)
[ ln(x + 2x2 + 1)
Other choices for <target> are diff, which is discussed in Section 11.5; and
exp, fact, heaviside, ln, sign, and sincos, which are obtained by choosing the
corresponding menu items. Many other target functions are available, but are not
on the menu; consult the Help for rewrite to see the list.
Functions Not on the Toolbar There are many useful functions that are not on
the toolbar. One example is the function collect(p,x). It collects coef cients of
like powers of x in the expression p.
[ > p2 : (x3)^23*x5
[ 3x (x 3)2 5
[ > collect(p2, x)
[ x 2 9x 15
If there is more than one variable, you can specify which one to collect.
[ > p3: (x3)^25*x(y-4)^3
[ 5x (x 3)2 (y 4)3
[ > collect(p3,x)
[ x 2 11x (y 4)3 9
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You can perform operations involving more than one expression. For example,
[ > collect(p2*p3,x)
[ x4 20x 3 ((y 4)3 122) x2 (9(y 4) 3 225) x 14(y 4)3 126
Another function not on the Command bar is the subs function. You can replace a variable x in an expression p with the variable y by typing subs(p, x y).
For example,
[ > p4:3*x^25*x9
[ 3x2 5x 9
[ > subs(p4,xx1)
[ 5x 3(x 1)2 14
Note that p4 has not changed:
[ > 2*p4
[ 6x2 10x 18
To obtain a new variable, you must give it a name:
[ >p5:subs(p4,xx1)
[ 5x 3(x 1)2 14
[ >2*p5
[ 10x 6(x + 1)2 28
The remaining items on the General Math menu are discussed in later
sections.
Test Your Understanding
T11.2–1 Given the expressions E1 x3 15x2 75x 125 and E2 (x 5)2 20x, use MuPAD to
(a) nd the product E1E2 and express it in its simplest form.
(b) nd the quotient E1兾E2 and express it in its simplest form.
(c) evaluate the sum E1 E2 at x 7.1 in symbolic form and in
numeric form.
(Answers: (a) (x 5)5, (b) x 5, (c) 13,671兾1000 in symbolic form,
13.6710 in numeric form)
11.3 Algebraic and Transcendental Equations
MuPAD can solve algebraic and transcendental equations as well as systems of
such equations. A transcendental equation is one that contains one or more transcendental functions, such as sin x, ex, or log x. The appropriate function to solve
such equations is the solve function.
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The solve Function
The function solve(p) solves a symbolic expression or equation (an expression
containing the equals sign ) represented by the expression p. If the expression
p does not contain an equals sign, MuPAD assumes that the expression is set to 0.
For example, to solve the equation 5x 25 0 using solve(p), type
[ > solve(5*x250)
[ {[x 5]}
or
[ > y : solve(5*x250)
[ {[x 5]}
[ >3*y
[ {3[y 5]}
The expression need not equal 0.
[ > solve(5*x-25)
[ {[x 5]}
Using the version solve(p,x), we obtain
[ > solve(5*x250,x)
[ {5}
or
[ > y : solve(5*x250,x)
[ {5}
MuPAD Libraries
MuPAD has hundreds of functions, which are organized into libraries. The Standard library is the most important MuPAD library. It has the most frequently used
functions such as solve, and simplify. Except for the functions of the Standard
library, library functions have a pre x separated with ::. For example, the solve
function in the Standard library solves for analytic solutions, whereas the solve
function in the Numeric library is called as numeric::solve(p).
Library functions can also be called by their short notation if you rst export
them with the use function. One library we will see soon is the numeric library;
it contains the function polyroots, which obtains the roots of polynomials numerically rather than symbolically. We can call this function by typing numeric::
polyroots. However, you can “export” this library’s functions from the library
into MuPAD by typing use(numeric). After exporting, you can type polyroots
instead of the long form numeric::polyroots, for example. To avoid confusion as
to whether a library function has been exported, in this chapter we will always use
the long form.
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There are many libraries, and you can obtain more information about them
and their functions by searching for libraries in the Help system. You can obtain
a list of the functions in a particular library, say, the numeric library, by typing
info(numeric).
Selecting Functions from the Command Bar
You can type functions directly in the input region of the notebook, or you can
select functions from the Command bar, which contains a subset of all the
MuPAD functions. The following choices appear when you choose Solve under
General Math from the Command bar:
The Solve Menu
Exact
Numeric
Linear System
Polynomial Diophantine Equation
Recurrences
ODE
Selecting Exact puts solve(#) in the input box. You then delete the # and
enter the equation to be solved. Instead of using the menu, you can type solve(p)
or solve(p,x), where p represents the equation to be solved and x is the solution
variable. To solve x2 4x 6 0, type
[ >solve(x^2-4*x6)
[ {[x = 2 - 12i], [x = 2 + 12i]}
Some nonlinear equations have an in nite number of solutions. MuP AD indicates these in the following way, if it can nd them exactly .
[ > solve(sin(2*x)-cos(x)0,x)
c e
2k
+ 2k ` k Z f h e +
` k Zf
2
6
3
The h symbol means that the complete solution can be obtained from the
expressions on both sides of the h symbol, where the k Z symbols indicate the
two solutions given are true for any integer value of k.
To nd all the roots of x4 1, you type solve(x^41) to get x 1 and x i. To nd only the real roots of x4 1, you can restrict the results to be positive
by using the option Type::Real as follows:
[ > assume (x, Type::Real): solve(x^4 1, x)
[ {1, 1}
Be sure to type unassume(x): afterward if you intend to use x later as another
variable. There are other options, such as Type::Positive, available to use with
assume and solve. We will see some of these later; see the Help for details.
You need not use the assume function. For example, there are an in nite
number of solutions to the equation e2x 3ex 54, and MuPAD can nd them
using solve. Suppose, however, that you are interested only in the real solutions.
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Then you can use the Real option as follows to obtain the only real solution,
x ln 6.
[ > solve(exp(2*x) 3*exp(x)54, x, Real)
[ {ln(6)}
You must always be careful in interpreting the results. For example, the equation x4 5x2 6x 2 0 is fourth order and therefore has four roots, but MuPAD
gives only three roots and does not indicate that the root x 1 is repeated.
[ > solve(x^4 - 5*x^2 6*x 2, x)
[ {1, - 13 - 1, 13 - 1}
You can avoid this problem by using the option Multiple, as follows:
[ > solve(x^4-5*x^2 6*x 2, x, Multiple)
[ {[1, 2], [1, - 13 - 1, 13 - 1]}
The result [1, 2] indicates that the root x 1 is repeated twice.
You can specify the interval over which you seek a solution. For example,
the equation cos(x兾5) 0 has an in nite number of solutions, which can be
found using the solve function. However, if you want only the solutions on the
interval 10 x 10, type
[ >solve(cos(x*PI兾5) 0, x -10..10)
c e-
15 5 5 15
,- , , f
2
2 2 2
MuPAD cannot always nd an exact solution. In such cases, MuP AD simply
displays the command. You can use the oat function to force MuPAD to produce a numerical solution. For example,
[ > solve(cos(x) x^2, x)
[ solve(cos(x) x2 0, x)
[ > oat(%)
[ 0.8241323123
You can also solve inequalities. For example, the inequality x2 3x > 4 has
solutions on the intervals - q 6 x 6 - 4 and 1 6 x 6 q .
[ > solve (x^2 3*x > 4, x)
[ (1, q ) h (- q ,- 4)
You can instruct MuPAD to make assumptions about the problem. For
example, the quadratic equation ax 2 bx c 0 has the following types of
solutions, depending on the values of the coef cients:
1. x = - b 2b - 4ac
if a
2a
c
2. x = if a 0 and b 0
b
2
0
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3. There is no solution if a b 0 but c
0.
4. There are an in nite number of solutions if a b c 0.
You can eliminate the last three cases by telling MuPAD to assume that a
as follows.
0,
[ > solve(a*x^2 b*x c, x, assume (a<>0))
c e-
b + 2 - 4ac + b2 b - 2 - 4ac + b2
f
,2a
2a
which is equivalent to the solution given previously. You can relax the assumption on a for later solutions by typing unassume(a): in the input box. Do not
forget to delete variables if you intend to reuse their names. The names x and y
are commonly used, so you should delete them before continuing.
Numeric
Selecting Numeric from the solve menu puts numeric::solve(#) in the input
box. The results are returned as oating-point numbers.
[ > numeric::solve(x^2-4*x6,x)
[ {2 1.414213562i, 2 1.414213562i}
Note that MuPAD might not be able to solve any given nonlinear equation,
and if so, you must use the numerical solver. However, in contrast to polynomial
equations, the numerical solver computes at most one solution of a nonpolynomial equation.
[ > numeric::solve(x2*exp(-x)-3)
[ {[x 0.583073876]}
Sometimes we can direct the search for a solution. Suppose we are interested
only in a solution over the range 0 x 10. The session is
[ > numeric::solve(x2*exp(-x)-3, x 0..10)
[ {[x 2.888703356]}
Sets of Equations
The solve function can also solve sets of equations. The solutions can be
expressed as a set of substitutions or as vectors. For example, to solve the set
x 2y 1, 3x 7y 5, you type
[ > solve ([x 2*y 1, 3*x 7*y 5], [x, y])
[ {[x 3, y 2]}
or
[ > solve([x 2*y 1, 3*x 7*y 5], [x, y], VectorFormat)
-3
c ea bf
2
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Consider another example:
[ > eqs : {x y a, x - a*y b}:
[ > solve(eqs, [x, y], IgnoreSpecialCases)
c e cx =
a2 + b
a - b
,y =
df
a + 1
a + 1
Here there are two symbolic parameters a and b, and thus we need to specify the symbols representing the variables to be found (here, x and y). The solution is incorrect if a 1, and the option IgnoreSpecialCases tells MuPAD to
ignore this possibility.
The solve function can also handle sets of nonlinear equations. For example,
[ > solve([x^2 y 1, x 2*y 1], [x, y])
1
3
c e [x = 1, y = 0], c x = - , y = d f
2
4
Test Your Understanding
T11.3–1 Use MuPAD to solve the equation 21 - x2 = x. (Answer: x = 12>2)
T11.3–2 Use MuPAD to solve the equation set x 6y a, 2x 3y 9 in terms
of the parameter a. (Answer: x (a 18)兾5, y (2a 9)兾15)
Here is a somewhat more involved application.
Intersection of Two Circles
EXAMPLE 11.3–1
We want to nd the intersection points of two circles. The rst circle has a radius of 2
and is centered at x 3, y 5. The second circle has a radius b and is centered at x 5,
y 3. See Figure 11.3–1.
(a) Find the (x, y) coordinates of the intersection points in terms of the parameter b.
(b) Evaluate the solution for the case where b = 13.
y
2
5
b
3
3
5
Figure 11.3–1 Intersection points of
two circles.
x
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485
■ Solution
(a) The intersection points are found from the solutions of the two equations for the
circles. These equations are
(x - 3)2 + (y - 5)2 = 4
for the rst circle and
(x - 5)2 + (y - 3)2 = b2
The session to solve these equations is the following.
[> p1:(x-3)^2 (y-5)^2-4
[ (x 3)2 (y 5)2 4
[ > p2:(x-5)^2 (y-3)^2-b^2
[ (x 5)2 (y 3)2 b2
[ > ans1:solve([p10,p20],[x,y])
c e cx =
b2
9
+
2
8
1,
y =
b2
+
8
1 +
7
d f e cx = 2
1 -
b2
9
+ ,
8
2
y = -
1 +
[ where
b4
3b2
1
A 16
2
J 1
2
These are the x and y coordinates of the two intersection points.
(b) Continue the above session by substituting b = 13 into the expression for x.
[ > ans2:subs(ans1,bsqrt(3))
c e cx =
147
147
147 33
31
33
, y =
d f e cx = + ,
8
8
8
8
8
8
[ > oat(ans2)
[ {[x 4.982,
y=
147 31
+
df
8
8
y 4.732]} {[x 3.268, y 3.018]}
These are the two pairs of coordinates for the two intersection points if b = 13.
Test Your Understanding
T11.3–3 Find the x and y coordinates of the two intersection points in Example 11.3–1 using b = 15. (Answer: x 4.986, y 5.236; x 2.764, y 3.014)
T11.3–4 Obtain the exact solution for x from the following equations.
x + 13y = e2
3x - y = 4
(Answer: x = (e2 + 4 13)>( + 3 13))
b2
7
+ df
8
2
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CHAPTER 11 MuPAD
Positioning a Robot Arm
EXAMPLE 11.3–2
Figure 11.3–2 shows a robot arm having two joints and two links. The angles of rotation
of the motors at the joints are 1 and 2. From trigonometry we can derive the following
expressions for the (x, y) coordinates of the hand.
x = L1 cos 1 + L2 cos( 1 +
2)
y = L1 sin 1 + L2 sin( 1 +
2)
Suppose that the link lengths are L1 4 ft and L2 3 ft.
Compute the motor angles required to position the hand at x 6 ft, y 2 ft.
■ Solution
Substituting the given values of L1, L2, x, and y into the above equations gives
6 = 4 cos 1 + 3 cos( 1 +
2)
2 = 4 sin 1 + 3 sin( 1 +
2)
The following session solves these equations. The variables th1 and th2 represent
and 2 .
1
[ >numeric::solve([4*cos(th1)3*cos(th1th2)6,4*sin(th1)3*sin(th1th2)2],[th1,th2])
[ {[th1 0.05756292117, th2 0.8956647939]}
Thus the solution in degress is
1 3.2981 ,
2 51.3178 .
y
Hand
L2
L1
θ2
Elbow Motor
θ1
x
Base Motor
Figure 11.3–2 A robot arm having two joints and
two links.
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11.3
Algebraic and Transcendental Equations
Selecting Linear System from the Solve menu inserts linsolve(#) into the
input box. This function solves sets of linear equations numerically rather than
symbolically.
Polynomial Diophantine Equations
The next item on the Solve menu is Polynomial Diophantine Equation. It
inserts the function solvelib::pdioe(#, #) into the input box. Such an equation is
one whose solutions are restricted to be polynomials. An important type of this
class has the form
au b␷ c
where a, b, and c are given polynomial functions of x, and u and ␷ are the
unknowns. This type of equation can be solved with the solve::pdioe(a,b,c,x)
function, for example,
(x2 + 1)u + (x3 + 1)␷ = 2x
The solution is found as follows.
[ > solvelib::pdioe((x^2 1), (x^3 1), 2*x, x)
[ x2 x 1, x 1
Thus one solution is u x 2 x 1 and ␷ x 1. Polynomial diophantine
equations do not have unique solutions. Any multiple wab can be used to transform
u and v into another solution u1, ␷1, where u1 u wb and ␷1 ␷ wa. For example, the choice w 1 leads to the solution u1 x3 x2 x, ␷1 x2 x.
Recurrence Relations
An example of a recurrence relation, sometimes called a difference equation, is
y(n 2) 2y (n 1) y (n) 5
Such equations can be solved using the menu item Recurrences on the Solve
menu. For example, with the given initial conditions y(0) 0, y(1) 1,
[ > eqn : y(n 2) - 2*y(n 1) y(n) 5:
[ > solve (rec(eqn, y(n), {y(0) 0, y(1) 1}))
c e
5n2
3n
f
2
2
Thus the solution is
y(n) =
5n2
3n
2
2
n = 2, 3, . . .
The last menu item under the Solve menu is ODE, which stands for the
ordinary differential equation solver. This is discussed in Section 11.6.
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CHAPTER 11 MuPAD
Optimization Problems
There are other solvers that deal with sets of linear inequalities, rather than linear equations. One class of such problems is called linear programming. A linear
programming problem is one in which you need to nd the values of a set of n
variables, denoted x1, x2, . . . , xn required to either maximize or minimize a function J(x1, x2, . . . , xn) subject to a set of equalities or inequalities that are linear
functions of the variables x1, x2, . . . , xn. Such problems have found widespread
applications in industry, including optimizing airline schedules and many types
of production problems. They can be solved using the functions in the linopt
library.
An example of such a problem is the following. Suppose we want to nd the
values of x, y, and z to maximize the function
J x y 2z
subject to the constraints
3x 4y 3z 20
6x 4y 3z 10
7x 4y 11z 30
In MuPAD we would type
[ > eq1 : 3*x 4*y - 3*z < 20
[ > eq2 : 6*x - 4*y - 3*z 10
[ > eq3 : 7*x 4*y 11*z < 30
[ > J : [{eql,eq2,eq3},- x y 2*z]
[ > linopt::maximize(J)
c c OPTIMAL, e x =
200
110 20
, y = 0, z =
f, d
87
87
87
The answer gives the values of x, y, and z that maximize J subject to the three
constraints. The resulting maximum possible value of J is given as 20兾87.
Sometimes we must specify constraints that are not immediately obvious. For
example, Example 8.3–4, Production Planning, has the constraint equations
5x 3y 3z 40
3x 3y 4z 30
where x, y, and z are the number of tons of each product to be produced each
week. The pro t to be maximized is
P 400x 600y 100z
Without additional constraints, the solver may give physically unrealistic values.
Here x, y, and z cannot be negative, so we add these three additional constraints.
The session is
[ > eqn1 : 5*x3*y3*z40
[ > eqn2 : 3*x3*y4*z30
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Linear Algebra
[ > eqn3 : x > 0: eqn4: y > 0: eqn5 : z > 0:
[ > J : [eqn1,eqn2,eqn3,eqn4,eqn5, 400*x600*y100*z]:
[ > linopt::maximize(J)
[ [OPTIMAL, {x 5, y 5, z 0}, 5000]
The maximum pro t is $5000.
Functions can be minimized with the linopt::minimize(J) function. The linopt
library has many more functions that are useful for linear programming problems.
11.4 Linear Algebra
The linalg library contains over 40 functions dealing with linear algebra. Type
info(linalg) to see a list of the functions. In this section we will use only a few of
these. A reminder: library functions can also be called by their short notation if
you rst export them with the use function, for example, use(linalg). After exporting, you can type charpoly instead of the long form linalg::charpoly, for
example. To avoid confusion as to whether a function has been exported, we will
always use the long form.
Matrix Operations
Linear algebra deals with matrices. The easiest way to create a matrix is the function matrix, which is in the Standard library. For example,
[ > A : matrix([[3, 5], [-4, 2]]),B : matrix([[2,1], [3, -2]])
3
5
2
1
冤 冢4 2冣, 冢3 2冣
Note that MuPAD surrounds matrices with a pair of parentheses, whereas
square brackets are used in MATLAB and much of the technical literature.
You can add, subtract, multiply, or divide matrices using the standard arithmetical operators. Adding or subtracting a scalar has the effect of adding or
subtracting that scalar from only the diagonal elements in the array. Note that
this is different from MATLAB, in which the scalar is subtracted from every
element.
[ > A * A, 1兾A, 2*A,A-3
5
1
13
26
- 11
25
6 10
0
≤, a
c a
b, ±
b, a
2
- 20 - 16
3
-8
4
-4
13
26
The determinant and rank are calculated as follows.
[ > linalg::det(A),linalg::rank(A)
[ 26, 2
5
b
-1
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CHAPTER 11 MuPAD
Matrix entries can be accessed with the index operator [ ].
[ > A[2,1]
[ 4
You can multiply matrices.
[ > A*B
c a
21
-2
-7
b
-8
You can create symbolic matrices.
[ > C : matrix([[a, b], [c, d]])
c a
a
c
b
b
d
Create a diagonal matrix as follows:
[ > D : matrix(3, 3, [2, -2, 5], Diagonal)
2
0
J P
0
0
-2
0
0
0
Q
5
Consider the rotation matrix R(a) for a coordinate system rotated through an
angle a:
[ > R : matrix([[cos(a),sin(a)],[-sin(a),cos(a)]])
cos(a) sin(a)
b
- sin(a) cos(a)
One property of coordinate rotations is that if we rotate the coordinate system
twice by the same angle to produce a third coordinate system, the result is the
same as a single rotation with twice the angle. Thus, R(a) R(a) should equal
R(2a). Let us check this with MuPAD as follows.
c a
[ > simplify(R*R)
cos(2a) sin(2a)
b
- sin(2a) cos(2a)
which is the same as R(2a):
[ > subs(R, a2*a)
c a
cos(2a) sin(2a)
b
- sin(2a) cos(2a)
To evaluate a symbolic matrix numerically, use the oat and subs functions.
[ > oat(subs(R, a 2))
- 0.4161468365
0.9092974268
b
ca
- 0.9092974268 - 0.4161468365
c a
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Linear Algebra
Note that this operation did not change the symbolic matrix R.
A rotation in the negative direction is represented by R(a), which equals
the inverse of R(a). The inverse of a matrix can be found from its reciprocal 1兾A
or with the inverse function, which is in the Standard library. You get the message FAIL if the matrix is not invertible.
[ > simplify(inverse(R))
c a
cos(a)
sin(a)
- sin(a)
b
cos(a)
This equals R(a), as can be shown by typing simplify(subs(R,aa)).
The Characteristic Polynomial, Eigenvalues, and Eigenvectors
The function linalg::charpoly(A, x) computes the characteristic polynomial of
the matrix A. Using the matrix A created previously, we have
[ > p : linalg::charpoly(A,x)
[ x2 5x 26
The eigenvalues are the roots of the characteristic polynomial and can be
found as follows.
[ > solve(p,x)
c e
179i 5
179i
5
, +
f
2
2 2
2
If you do not need the characteristic polynomial, you can calculate the eigenvalues of A directly with the function linalg::eigenvalues(A). This gives the
exact solution for the eigenvalues. This might not be possible for large matrices;
if so, you can use the numeric::eigenvalues(A) function.
The function linalg::eigenvectors(A) computes both the eigenvalues and
the eigenvectors of the matrix. The function returns a list of sublists; each sublist
consists of an eigenvalue of A, its algebraic multiplicity, and a basis vector,
which is an eigenvector normalized so that the last element is 1. If a basis vector
cannot be computed, the message FAIL is returned. For example,
[ > A : matrix([[3, 5], [-4, 2]]):
[ > linalg::eigenvectors(A)
1
179i
1
179i
5
179i
5
179i
- +
- , 1,
+
,
1,
,
8
8
8
8
2
2
J JJ2
JP
JP
QKK J2
QKKK
1
1
Matrix Solution of Linear Equations
The function linalg::matlinsolve(A, b) computes the general solution of the
equation Ax b, where A is an m n matrix, b is an m 1 vector, and the solution x is an n 1 vector. Consider the set
2x y 11
3x 5y 16
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It can be solved as follows.
[ > A : matrix([[2,1], [3, -5 ]]):b: matrix ([11, 16]):
[ > x : linalg::matlinsolve(A, b)
3
c a b
5
This set has a unique solution x 3, y 5, whereas the set
2x 3y 14
4x 6y 10
has no solution. This is indicated by an empty set of brackets as the result.
[ > A : matrix ([[2,3], [4,6]]): b : matrix([14, 10]):
[ > x : linalg:: matlinsolve(A, b)
[ []
For this set of equations the rank of A equals 1, which is less than the rank of the
matrix [A, b], which is created by typing A.b. Thus there is no solution. These
ranks can be computed as follows.
[ > linalg::rank(A)
[ 1
[ > C : A.b:
[ > linalg::rank(C)
[ 2
See Chapter 8 for discussion of the existence and uniqueness of solutions of linear algebraic equations.
Now consider the following set, which has the same matrix A but an in nite
number of solutions, x 7 3y兾2.
2x 3y 14
4x 6y 28
The MuPAD session is
[ > A : matrix([[2,3], [4,6]]); b : matrix([14, 28]):
[ > C: A.b:
[ > linalg::rank(A)
[ 1
[ > linalg::rank(C)
[ 1
[ > x : linalg::matlinsolve(A, b)
3
7
a b,
2
J J 0 JP
QKK
1
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Calculus
Because the rank of A equals the rank of [A, b], but is less than the number of
equations, there are an in nite number of solutions. MuP AD expresses this fact
indirectly in vector form. The rst vector on the solution, (7, 0), says that one solution of the set is y 0 and x 7. The square brackets enclosing the second
vector (3兾2, 1) may be thought of as an eigenvector that expresses the slope of
x versus y in the solution x 7 3y兾2.
Equation sets can be solved symbolically. Consider the set
2x ay 1
3x 6y 3
It can be solved as follows.
[ > A:matrix([[2,a],[3,-6]])
[ > b : matrix([[1],[3]])
[ > linalg::matlinsolve(A,b)
6a +12
6a +24
≥ ±
6 ≤
6a+ 24
Use the function numeric::matlinsolve to solve a linear system numerically.
Test Your Understanding
T11.4–1 Consider three successive coordinate rotations using the same angle a.
Show that the product RRR of the rotation matrix R(a) equals R(3a).
T11.4–2 Find the characteristic polynomial and roots of the following matrix.
A = c
-2
-3k
1
d
-5
(Answers: s2 7s 10 3k and s = (- 7 19 - 12k)>2)
T11.4–3 Use the matrix inverse and the matlinsolve function to solve the following set.
4x 6y 2
7x 4y 23
(Answer: x 5, y 3)
11.5 Calculus
In Chapter 9 we discussed techniques for performing numerical differentiation and
numerical integration; this section covers differentiation and integration of symbolic expressions to obtain closed-form results for the derivatives and integrals.
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Differentiation
The diff(f,x) function is used to obtain the symbolic ordinary derivative of the
expression f with respect to x. Although this function has the same name as the
function used to compute numerical differences (see Chapter 9), MuPAD uses
the symbolic form. The differential operator D(f), abbreviated as f, computes
the derivative of the univariate function f. The D operator is used for functions,
not expressions. We will not use it here.
For example, the derivatives
d(x n)
= nxn - 1
dx
d(ln x)
1
=
x
dx
2
d(sin x)
= 2 sin x cos x
dx
d(sin y)
= cos y
dy
[sin (xy)]
= y cos(xy)
x
are obtained with the following session.
[ > diff(xn, x)
[ nxn1
[ > diff(ln(x),x)
1
c
x
[ > diff(sin(x)2,x)
[ 2 cos (x) sin(x)
[ > diff(sin(y),y)
[ cos(y)
When there is more than one variable, the diff function computes the partial
derivative. For example, if
f(x, y) sin(xy)
then
f
= y cos (xy)
x
The corresponding session is
[ > diff(sin(x*y),x)
[ y cos(xy)
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495
Higher derivatives can be obtained by successive application of the diff
command. For example,
[ > diff(diff(ln(x),x),x)
1
c - 2
x
A shorter form is
[ > diff(ln(x),x,x)
The mixed second partial derivative may be calculated as follows:
[ > diff((x2*y3 6*x2*y), y, x)
[ 6xy2 12x
Max-Min Problems
The derivative can be used to nd the maximum or minimum of a continuous
function, say f(x), over an interval a x b. A local maximum or local minimum
(one that does not occur at one of the boundaries x a or x b) can occur only
at a critical point, which is a point where either df兾dx 0 or df兾dx does not exist.
If d 2f兾dx 2 0, the point is a relative minimum; if d 2f兾dx2 0, the point is a relative maximum. If d 2f兾dx2 0, the point is neither a minimum nor a maximum,
but is an in ection point . If multiple candidates exist, you must evaluate the function at each point to determine the global maximum and global minimum.
Topping the Green Monster
EXAMPLE 11.5–1
The “Green Monster” is a wall 37 ft high in left eld at Fenway Park in Boston. The wall
is 310 ft from home plate down the left eld line. Assuming that the batter hits the ball
4 ft above the ground, and neglecting air resistance, determine the minimum speed the
batter must give to the ball in order to hit it over the Green Monster. In addition, nd the
angle at which the ball must be hit. (see Figure 11.5–1).
■ Solution
The equations of motion for a projectile launched with a speed ␷0 at an angle relative to
the horizontal are
gt2
+ (␷0 sin )t
2
where x 0, y 0 is the location of the ball when it is hit. Because we are not concerned
with the time of ight in this problem, we can eliminate t and obtain an equation for y in
terms of x. To do this, solve the x equation for t and substitute this into the y equation to
obtain
g x2(t)
y(t) = - 2
+ x(t) tan
2 ␷ 0 cos2
x(t) = (␷0 cos )t
y(t) = -
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CHAPTER 11 MuPAD
100
90
80
70
Height y (ft)
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60
50
40
30
Green
Monster
20
10
0
0
50
100
150
200
Distance x (ft)
250
300
350
Figure 11.5–1 A baseball trajectory to clear the Green Monster.
(You could use MuPAD to do this algebra if you wish. We will use MuPAD to do the more
dif cult task to follow .)
Because the ball is hit 4 ft above the ground, the ball must rise 37 4 33 ft to clear
the wall. Let h represent the relative height of the wall (33 ft). Let d represent the distance to
the wall (310 ft). Use g 32.2 ft/sec2. When x d, y h. Thus the previous equation gives
h = -
g
d2
2
2 ␷0 cos2
+ d tan
which can easily be solved for ␷20 as follows.
␷20 =
g
d2
2 cos2 (d tan
- h)
Because ␷0 0, minimizing ␷20 is equivalent to minimizing ␷0. Note also that gd 2兾2 is a
multiplicative factor in the expression for ␷20. Thus the minimizing value of is independent of g and can be found by minimizing the function
f =
1
2
cos (d tan
- h)
The session to do this is as follows. The variable th represents the angle of the ball’s
velocity vector relative to the horizontal. The rst step is to calculate the derivative
df兾d and solve the equation df兾d 0 for . We know must be less than /2 L
1.57 rad.
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[ > d : 310:h:33:g:32.2:
[ > f : 1兾(((cos(th))^2)*(d*tan(th)-h)):
[ > dfdth : diff(f, th):
[ > thmin : numeric::realroot(dfdth,th 0..1.57)
[ 0.838424752
So the solution candidate is 0.8384 rad, or about 48°. To verify that this is a minimum
solution, and not a maximum or an in ection point, we can check the second derivative
d2f兾d 2. If this derivative is positive, the solution represents a minimum. To check this
and to nd the speed required, continue the session as follows.
[ > second : diff(diff(f,th),th):
[ > subs(second, th thmin)
[ > oat(%)
[ 0.03209697518
So the second derivative is positive, and the solution represents the minimum. To nd the
speed required, continue the session as follows.
[ > v2:(g*d^2兾2)*f
[ > v2min : subs(v2, th thmin):
[ > vmin :sqrt(v2min)
[ 105.3612757
Thus the minimum speed required is about 105 ft/sec, or about 72 mph. A ball hit with this
speed will clear the wall only if it is hit at an angle of approximately 48°.
Test Your Understanding
T11.5–1 Given that y sinh (3x) cosh(5x), use MuPAD to nd dy兾dx at x 0.2.
(Answer: 9.2288)
T11.5–2 Given that z 5 cos (2x) ln(4y), use MuPAD to nd z兾y. (Answer:
5 cos(2x)兾y)
Integration
The int(f,x) function is used to integrate a symbolic expression f with respect
to x. It is possible that the integral does not exist in closed form, or that MuPAD
cannot nd the integral even if it exists. For example, you can obtain the following integrals with the session shown below.
1
dx = ln x
Lx
cos x dx = sin x
L
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L
sin y dy = - cos y
L
xn dx =
xn + 1
n + 1
[ > int(1兾x,x)
[ ln(x)
[ > int(cos(x),x)
[ sin(x)
[ > int(sin(y),y)
[ cos(y)
[ > assume(n <>-1):int(xn,x)
xn+1
c
n+1
If we had not forced MuPAD into assuming that n
1, MuPAD would have
given us a second solution, ln x, as well.
De nite integrals can be computed by giving the limits after the integrand.
For example, the integral
10
x3dx
L0
is computed as follows:
[ > int(x3, x0..10)
[ 2500
The following session gives an example for which no integral can be found.
The inde nite integral exists, but the de nite integral does not exist if the limits
of integration include the singularity at x 1. The integral is
The session is
[ > int(1兾(x-1),x)
[ ln(x 1)
[ > int(1兾(x-1),x0..2
[ unde ned
1
dx = ln ⱍ x - 1ⱍ
x
1
L
Test Your Understanding
T11.5–3 Given that y x sin(3x), use MuPAD to nd 1 ydx. (Answer: (sin(3x) 3x cos(3x))兾9)
T11.5–4 Given that z 6y2 tan(8x), use MuPAD to nd 1 zdy. (Answer:
2y3 tan(8x))
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11.5
Calculus
T11.5–5 Use MuPAD to evaluate
5
L- 2
x sin(3x) dx
(Answer: 0.6672)
Taylor Series
The taylor(f,xa,n) function gives the rst n 1 terms in the Taylor series for
the function de ned in the expression f, evaluated at the point x a. The parameter n is optional. If the parameter a is omitted, the function returns the series
evaluated at x 0. Some common examples of the Taylor series are
ex = 1 + x +
x2
x3
x4
+
+
+ Á
2!
3!
4!
-q 6 x 6 q
sin x = x -
x5
x7
x3
+
+ Á
3!
5!
7!
-q 6 x 6 q
where a 0 in both examples.
Here are some examples.
[ > taylor(exp(x), x)
c 1 + x +
x3
x4
x5
x2
+
+
+
+ O(x6)
2
6
24
120
[ > taylor(exp(x), x, 8)
c 1 + x +
x2
x3
x4
x5
x6
x7
+
+
+
+
+
+ O(x8)
2
6
24
120
720
5040
[ > taylor(sin(x),x,6)
c x -
x5
x3
+
+ O(x7)
6
120
A linear approximation to ex about the point x 2 is found as follows.
[ > taylor(exp(x),x2,2)
[ e2 e2(x 2) O((x 2)2)
The latter expression corresponds to
ex L e2 [1 (x 2)]
for x L 2
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Sums
The sum(f,k a..b) function returns the sum of the expression f as the symbolic
variable varies from a to b. Here are some examples. The summations
10
ak = 0 + 1 + 2 + 3 +
Á + 9 + 10 = 55
n-1
Á + n - 1 = n(n - 1)
2
k=0
ak = 0 + 1 + 2 + 3 +
k=0
4
2
a k = 1 + 4 + 9 + 16 = 30
k=1
are given by
[ > sum(k,k 0..10)
[ 55
[ > sum(k, k 0..n-1)
n(n - 1)
c
2
[ > sum(k2, k 1..4)
[ 30
Limits
The function limit(f,x a) returns the limit
lim f(x)
x:a
For example, the limits
lim
x - 3
x : 3 x2 - 9
=
1
6
sin(x + h) - sin(x)
= cos x
x:0
h
lim
are found from
[ > limit((x-3)兾(x^2-9), x3)
1
c
6
[ > limit((sin(x h)-sin(x))兾h,h0)
[ cos(x)
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Ordinary Differential Equations
The forms limit(f, x a, Left) and limit(f, x a, Right) specify the direction of the limit. For example,
lim
1
x : 0- x
lim
1
x : 0+ x
= -q
= q
are given by
[ > limit(1兾x, x0, Left)
[ -q
[ > limit(1兾x, x0, Right)
[ q
Test Your Understanding
T11.5–6 Use MuPAD to nd the rst three nonzero terms in the Taylor series for
cos x.
(Answer: 1 - x2>2 + x4>24)
T11.5–7 Use MuPAD to nd a formula for the sum
m-1
3
am
m=0
(Answer: m4>4 - m3>2 + m2>4)
T11.5–8 Use MuPAD to evaluate
7
a cos(n)
n=0
(Answer: 0)
T11.5–9 Use MuPAD to evaluate
lim
2x - 10
x : 5 x3 - 125
(Answer: 2>75)
11.6 Ordinary Differential Equations
Methods for obtaining a numerical solution to differentiation, integration, and
differential equation problems were covered in Chapter 9. However, we prefer to
obtain an analytical solution whenever possible, because it is more general, and
thus more useful for designing engineering devices or processes.
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MuPAD provides the ode and solve functions for solving ordinary differential equations in closed form. These functions are different from the numerical
ODE solvers in MATLAB (such as ode45 and the symbolic ODE solver
dsolve in the Symbolic Math toolbox).
The syntax of the MuPAD ode and solve functions differs somewhat according
to whether they are used to solve single equations or sets of equations, and whether
boundary conditions are speci ed. MuP AD also has the numeric::solve function
for solving differential equations numerically, but we will not cover this topic
because the MATLAB ode solvers are suf cient for our purposes.
Solving a Single Differential Equation
The MuPAD ode function is used to describe the equation and its boundary conditions, if any, but it does not solve the equation. The solve function is then used to
solve the problem. The arbitrary constants in the solution are denoted by Cl, C2, and
so on. The number of such constants is the same as the order of the equation. Derivatives are denoted by primes regardless of the independent variable. For example,
y¿ =
dy
dy
=
dx
dt
y– =
d2y
d2y
=
dx2
dt2
and so on.
For example, the equation
dy
+ 2y = 12
dt
(11.6–1)
has the solution
y(t) = 6 + C1e - 2t
The solution can be found with the following session. We give the equation
the arbitrary name eqn and tell MuPAD that the independent variable is t and the
dependent variable is y(t). If the colon is not used after the command, MuPAD
will display the equation.
[ > eqn : ode(y¿(t)2*y(t)12, y(t)):
[ > solve (eqn)
C1
c e 2t + 6 f
e
which is the correct solution, although it is expressed in a less than conventional
form.
Suppose the initial condition is y(0) 5. We nd the solution as follows.
[ > eqn : ode ({y¿(t)2*y(t)12, y(0)5}, y(t)):
[ > solve(eqn)
1
c e 6 - 2t f
e
Note how the braces are used to group the equation and the initial condition.
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Ordinary Differential Equations
You need not use t as the independent variable or y as the dependent variable.
We could just as easily express Eq. (11.6–1) as
dw
+ 2w = 12
dx
and the MuPAD session would be
[ > eqn : ode(w¿(x)2*w(x)12, w(x)):
[ > solve(eqn)
C1
c e 2x + 6 f
e
There can be symbolic constants in the equation and boundary conditions,
although the more complicated the equation, the more complicated the solution
will be, perhaps to the point of not being useful. Here is a second-order example.
d2y
2
dt
= c2y
The session is
[ > eqn : ode(y¿¿(t)c^2*y(t), y(t)):
[ > solve(eqn)
C1
c e ct + C2ect f
e
The solution is y(t) = C1e-ct + C2ect .
With the initial conditions y(0) = 1, y¿(0) = 0, the session is
[ > eqn : ode({y¿¿(t)c^2*y(t), y(0)1, y¿(0)0}, y(t)):
[ > solve(eqn)
The solution given by MuPAD is y(t) = 1 if c = 0 and
y(t) =
if c
1 ct
1
e + e-ct
2
2
0.
Here is an example having a forcing function.
dy
= sin bt
dt
y(0) = 0
The session is
[ > eqn : ode({y¿(t)sin(b*t), y(0)0}, y(t)):
[ > solve(eqn)
cos(bt)
1
f
c e b
b
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The solution is
y(t) =
1
(1 - cos bt)
b
General boundary conditions may also be speci ed. For example, consider
the problem
d2y
dt2
+ 9y = 0
y(0) = 1,
.
y () = 2
The session to nd the solution and plot it is
[ > eqn : ode({y¿¿(t) 9*y(t)0, y(0)1, y¿(PI)2}, y(t)):
[ > solve(eqn)
c e cos(3t) -
2 sin(3t)
f
3
[ > plot( oat(%))
So the solution is
y(t) = cos 3t -
2
sin 3t
3
If a plot of the solution is required, then it is better to make use of the powerful
MATLAB plot function.
Higher-order equations can also be solved. Consider, for example,
d 4y
dt4
- y = 0
with the initial conditions y(0) = 1 and
dy
d2y
d3y
`
= 2`
= 3`
= 0
dt t = 0
dt t = 0
dt t = 0
The session is
[ > eqn : ode({y¿¿¿¿(t)-y(t)0, y(0)1, y¿(0)0, y¿¿(0)0, y¿¿¿(0)0}, y(t)):
[ > solve(eqn)
c e
cos(t)
1
et
+
+ f
t
2
4
4e
So the solution is
y(t) =
1
1
1 -t
e + cos t + et
4
2
4
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Ordinary Differential Equations
Solving Sets of Equations
Sets of equations can be solved with solve. Use a set of square brackets to denote
the set of equations and another set of brackets to denote the dependent variables.
For example, consider the set
dx
= 3x + 4y
dt
dy
= - 4x + 3y
dt
The MuPAD session is
[ > eqn : ode([x¿(t)3*x(t)4*y(t), y¿(t) 4*x(t)3*y(t)], [x(t),y(t)]):
[ > solve(eqn)
[ {[y(t) = -C1 cos(4t)e3t - C2 sin(4t)e3t, x(t) = C2 cos(4t)e3t - C1 sin(4t)e3t]}
Thus the solution is
y(t) = - C1e3t cos 4t - C2e3t sin 4t
x(t) = C2e3t cos 4t - C1e3t sin 4t
Note that MuPAD rst displays the solution for the second variable y(t).
Sets of equations with speci ed boundary conditions can be solved as follows. For example, consider the equation set just given, with the initial conditions x(0) 0, y(0) 1. The session is
[ > eqn:ode({x¿(t)3*x(t)4*y(t),y¿(t)4*x(t)3*y(t),x(0)0, y(0)1} [x(t),y(t)]):
[ {y(t) = cos(4t)e3t, x(t) = sin(4t)e3t}
Note how when boundary conditions are speci ed, braces are used instead of
brackets to group the equations and the boundary conditions.
Solving Nonlinear Equations
MuPAD can solve some nonlinear differential equations. For example, the problem
dy
= y2
dt
y(0) = 1
can be solved with the following session.
[ > eqn: ode({y¿(t)(y(t))2, y(0)1}, y(t))):
[ > solve(eqn)
1
f
c et - 1
(11.6–2)
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CHAPTER 11 MuPAD
So the solution is
1
1 - t
y(t) =
Not all nonlinear equations can be solved in closed form. An example is the
following equation, which is the equation of motion of a pendulum:
L " + g sin = 0. If you try to solve this equation in MuPAD, the result is expressed in terms of integrals that would need to be evaluated numerically.
Test Your Understanding
T11.6–1 Use MuPAD to solve the problem
d2y
2
dt
+ ay = 0
y(0) = 1
y. (0) = 0
for a 7 0. Check the answer by hand or with MuPAD. (Answer:
y(t) = cos( 1at))
11.7 Laplace Transforms
The Laplace transform L[y(t)] of a function y(t) is de ned to be
q
L[y(t)] =
L0
y(t)e-st dt = Y(s)
(11.7–1)
and can be obtained in MuPAD by typing transform::laplace(y,t,s), where y is
the function of t. The result is a function of s. The laplace command is in the
Transform library, which also contains the Fourier and z transforms.
Here is a MuPAD session with some examples. The functions are t3, ebt,
and sin bt.
[ > transform::laplace(t^3, t, s)
6
c 4
s
[ > transform::laplace(exp(b*t), t, s)
c
1
b + s
[ > transform::laplace(sin(b*t), t, s)
c
b
2
b + s2
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11.7
Laplace Transforms
The inverse Laplace transform L-1[Y(s)] is that time function y(t) whose
transform is Y(s), that is, y(t) = L - 1[Y(s)]. Inverse transforms can be found
using the function transform::invlaplace(Y, s, t). Note the reversed order of s
and t. For example,
[ > transform::invlaplace(6兾s^4, s, t)
[ t3
[ > transform::invlaplace(1兾(sb), s, t)
1
c bt
e
Depending on what computations have preceded the command in the notebook, you may get different, but still correct, answers. For example, sometimes,
but not always, you will get an unreduced answer like the following.
[ > transform::invlaplace(b兾(s^2b^2), s, t)
[ sin(2b2 t)
This occurs because MuPAD considers the possibility that b may be negative.
Note that MuPAD does not reduce 2b2 to b. If in fact b 7 0, you can avoid this
by typing
[ > assume(b>0):
[ > transform::invlaplace(b兾(s^2b^2), s, t)
[sin(bt)
Because the transform is an integral, its has the properties of integrals. In
particular, it has the linearity property, which states that if a and b are not functions of t, then
L[af1(t) + bf2(t)] = aL[f1(t)] + bL[f2(t)]
(11.7–2)
The transforms of derivatives are useful for solving differential equations.
Applying integration by parts to the de nition of the transform, we obtain
La
q
q
q
dy -st
dy
b =
e dt = y(t)e-st ` + s
y(t)e-st dt
dt
0
L0
L0 dt
= sL[y(t)] - y(0) = sY(s) - y(0)
(11.7–3)
This procedure can be extended to higher derivatives. For example, the result for
the second derivative is
La
d2y
dt2
#
b = s2Y(s) - sy(0) - y(0)
(11.7–4)
The general result for any order derivative is
La
n
dny
n
n-k
b
=
s
Y(s)
a s gk - 1
dtn
k=1
(11.7–5)
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CHAPTER 11 MuPAD
where
gk - 1 =
dk - 1y
`
dtk - 1 t = 0
(11.7–6)
Application to Differential Equations
The derivative and linearity properties can be used to solve equations such as
#
a y + y = b␷(t)
(11.7–7)
Application of the transform gives
Y(s) =
ay(0)
b
+
V(s)
as + 1
as + 1
(11.7–8)
The free response is given by
L-1c
ay(0)
y(0)
d = L-1c
d = y(0)e-t>a
as + 1
s + 1>a
The forced response is given by
L-1c
b
V(s) d
as + 1
(11.7–9)
This cannot be evaluated until V(s) is speci ed.
Suppose ␷(t) is a unit-step function, which is also called the Heaviside function. In MuPAD this is called by the command heaviside(t). For example,
[ > transform::laplace(heaviside(t), t, s)
1
c
s
Thus the transform of the unit-step function is 1兾s, and so V(s) 1兾s. Then Equation (11.7–9) becomes
L-1c
b
d
s(as + 1)
To nd the inverse transform, which is y(t), enter
[ > transform::invlaplace(b兾(s*(a*s1)), s, t)
cb -
b
t
ea
which can be expressed as b(1 - e-t>a). This is the forced response to a unit-step
input.
Consider the second-order model
$
#
x + 2 x + x = f(t)
(11.7–10)
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11.7
Laplace Transforms
Transforming this equation gives
#
[s2X(s) - sx(0) - x(0)] + 2[sX(s) - x(0)] + X(s) = F(s)
Solve for X(s).
X(s) =
#
x(0)s + x(0) + 2x(0)
F(s)
+
2
s + 2s + 1
2
s + 2s + 1
The free response is obtained from
#
x (0)s + x (0) + 2x(0)
d
s2 + 2s + 1
#
Suppose the initial conditions are x(0) = 2 and x (0) = 3. Then the free response
is obtained from
x(t) = L - 1 c
2s + 7
d
x(t) = L - 1 c 2
s + 2s + 1
(11.7–11)
It can be found in MuPAD by typing
[ > transform::invlaplace((2*s7)兾(s^22*s1), s, t)
2
5t
c t + t
e
e
So the free response is
x(t) = 2e-t + 5te-t
The forced response is obtained from
F(s)
x(t) = L - 1 c 2
d
s + 2s + 1
If f(t) is a unit-step function, F(s) = 1>s, and the forced response is
x(t) = L - 1 c
1
2
s(s + 2s + 1)
d
To nd the forced response, enter
[ > transform::invlaplace(1兾(s*(s^22*s1)),s,t)
t
1
c 1 - t - t
e
e
So the forced response is
x(t) = 1 - te-t - e-t
(11.7–12)
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CHAPTER 11 MuPAD
y
x
x
y
k
c
k
m
m
c
( a)
(b)
Figure 11.7–1 Two mechanical systems. The model for (a) contains the derivative
of the input y(t); the model for (b) does not.
Input Derivatives
Two similar mechanical systems are shown in Figure 11.7–1. In both cases the
input is a displacement y(t). Their equations of motion are
$
#
#
m x + cx + kx = ky + c y
(11.7–13)
$
#
m x + cx + kx = ky
(11.7–14)
The only difference between these systems is that the system in Figure 11.7–1a
has an equation of motion containing the derivative of the input function y(t).
Both systems are examples of the more general differential equation
$
#
#
m x + cx + kx = dy + gy
(11.7–15)
We now demonstrate how to use the Laplace transform to nd the step response of equations containing derivatives of the input. Suppose the initial conditions are zero. Then transforming Equation (11.7–15) gives
d + gs
Y(s)
(11.7–16)
X(s) =
2
ms + cs + k
Let us compare the unit-step response of Equation (11.7–16) for two cases
using the values m = 1, c = 2, and k = d = 1, with zero initial conditions. The
two cases are g = 0 and g = 5.
With Y(s) = 1>s, Equation (11.7–16) gives
X(s) =
1 + gs
2
s(s + 2s + 1)
(11.7–17)
The response for the case g = 0 was found earlier in Equation (11.7–14). The response for g = 5 is found by typing
[ > transform::invlaplace((15s*)兾(s*(s22*s1)), s, t)
c
4t
1
- t + 1
t
e
e
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11.7
Laplace Transforms
2.5
g=5
2
x
1.5
1
g=0
0.5
0
0
1
2
3
4
5
6
7
t
.
.
Figure 11.7–2 The step response of the model ẍ + 2x + x = u + gu for g = 0 and g = 5.
or
x(t) = 4te-t - e-t + 1
The two response are plotted in Figure 11.7–2. The effect of differentiating
the input is an increase in the response’s peak value.
Partial fraction expansions are useful when dealing with inverse Laplace
transforms and for other applications. The partfrac(f,x) function creates the partial fraction expansion of the expression f(s) in terms of its variable s.
[ > partfrac((6*s5)兾(s^314*s^259*s70),s)
c
25
7
37
6(s + 5)
15(s + 2)
10(s + 7)
Test Your Understanding
T11.7–1 Find the Laplace transform of the functions 1 - e-at and cos bt. Use the
ilaplace function to check your answers.
$
#
T11.7–2 Use the Laplace transform to solve the problem 5 y + 20y +
#
15y = 30u - 4u, where u(t) is a unit-step function and y(0) = 5,
#
y(0) = 1. (Answer: y(t) = - 1.6e-3t + 4.6e-t + 2)
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CHAPTER 11 MuPAD
11.8 Special Functions
Differential equations that do not have closed-form solutions can often be solved
in terms of special functions. Many such ordinary differential equations arise in
the solution of partial differential equations. Examples include the Chebyshev
polynomials of the rst kind T(n, x), the Hermite polynomials Hn(x), and several
types of Bessel functions. Other examples are the Airy functions Ai(x) and Bi(x),
which are the two independent solutions to Airy’s equation y⬘⬘ ⫺ xy ⫽ 0.
Table 11.8–1 lists some of the special functions available in MuPAD.
Consider a speci c case of Legendre’ s equation (1 ⫺ x2)y⬘⬘ ⫺ 2xy⬘ ⫹ 6y ⫽ 0
with the initial conditions y(0) ⫽ 1, y⬘(0) ⫽ 0. It can be solved in MuPAD with
the following session:
[ > eqn1:= ode({(1-x^2)*y¿¿ (x)-2*x*y¿ (x)+6*y(x)=0, y(0)=1, y¿(0)=0}, y(x)):
[ > solve(eqn1)
[ {1 ⫺ 3x2}
The result is a nite polynomial of degree 2.
Airy’s equation y⬘⬘ ⫺ xy ⫽ 0 with the same initial conditions can be solved
with the following session:
[ > eqn2:= ode({y¿¿ (x)-x*y(x)=0, y(0)=1, y¿(0) = 0}, y(x)):
[ > solve(eqn2)
2
2
31>6⌫ a b airyBi(x,0)
32>3⌫a b airyAi(x,0)
3
3
+
M
JL
2
2
where ⌫(x) is another special function called the gamma function. The gamma
function is evaluated by typing gamma(x). Table 11.8–2 shows that the oat
function is used to evaluate special functions numerically. For example,
[ > oat(gamma(2 兾3))
[ 1.354117939
[ > oat(airy Ai(2,0))
[ 0.03492413042
Table 11.8–1 Special function calls in MuPAD
Name and Symbol
Function Call
Airy, Ai(x)
Airy, Bi(x)
Chebyshev of rst kind, T(n, x)
Gamma, ⌫(x)
Hermite, Hn(x)
Bessel I, In(x)
Bessel J, Jn(x)
Bessel K, Kn(x)
Bessel Y, Yn(x)
Laguerre, L(n, a, x)
Legendre, Pn(x)
airy Ai(x)
airy Bi(x)
chebyshev1 (n, x)
gamma(x)
hermite (n,x)
besselI(n,x)
besselJ(n,x)
besselK(n,x)
besselY (n,x)
laguerreL(n,a,x)
legendre(n,x)
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11.8
Special Functions
Table 11.8–2 Evaluation of special functions in MuPAD
Result
Code
Symbolic nite series
Symbolic in nite series
Numeric result
orthpoly::
series
oat
If an explicit series solution is preferred, you can use the series option as follows. Using the particular Legendre’s equation (1 x2)y 2xy 12y 0 as
an example, solving it with the solve function, even with simple initial conditions such as y(0) 1 and y(0) 0, results in a complicated expression that is
too detailed to display here (try it). However, even with arbitrary initial conditions, the series solution is not very complicated. The session is
[ > eqn3 := (1-x^2)*y"(x)-2*y¿(x)+12*y(x)=0:
[ > ode::series({eqn3, y(0)=a, y¿0)=b},y(x), x=0)
5b 3
x + 3ax4 + O(x6) f
3
To evaluate a special function symbolically as a nite series , use either the
orthpoly:: option or the series function. The orthpoly package provides some
standard orthogonal polynomials. Call the package functions by using the package name orthpoly and the name of the function. For example, the fth-order
Legendre polynomial is obtained as follows:
c e a + bx - 6ax2 -
[ > orthpoly::legendre(5, x)
c e poly a
35x3
15x
63x5
+
, [x] b f
8
4
8
[ > orthpoly::legendre(5,2)
c
743
4
[ > oat (orthpoly::legendr e(5,2))
[ 185.75
To evaluate symbolically a special function that consists of an in nite series ,
use the series function. For example,
[ > series(besselJ(0,x), x)
c1 -
x2
x4
+
+ O(x6)
4
64
You can plot special functions just as any other functions. For example,
[ > plot(airyAi(x),x=0..5)
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CHAPTER 11 MuPAD
You can plot the solution of a differential equation by following this example based on Airy’s equation.
[ > eqn4 := ode({y"(x)=x*y(x), y(0)=1, y¿(0)=-0.1}, y(x)):
[ > solve(eqn4):
[ > y4 := op(%):
[ > plotfunc2d(y4, x=0..1)
The op function removes the braces surrounding the solution obtained by the
solve function.
Test Your Understanding
T11.8–1 Solve Legendre’s equation for the given initial conditions.
y(0) 12
y(0) 0
(1 x2)y 2xy 6y 0
2
(Answer: y(x) 12 36x )
T11.8–2 Evaluate the solution to Airy’s equation at x 3 for the given initial
conditions.
y xy 0
y(0) 1
y(0) 5
(Answer: y(3) 89.6423632)
11.9 Summary
This chapter covers a subset of the capabilities of MuPAD. Now that you have
nished this chapter , you should be able to use MuPAD to do the following:
■
■
■
■
■
■
■
■
Create symbolic expressions and manipulate them algebraically.
Obtain symbolic solutions to algebraic and transcendental equations.
Perform symbolic differentiation and integration.
Evaluate limits and series symbolically.
Obtain symbolic solutions to ordinary differential equations.
Obtain and apply Laplace transforms.
Perform symbolic linear algebra operations, including obtaining expressions for determinants, matrix inverses, eigenvectors, and eigenvalues.
Evaluate the special functions of mathematics.
Key Terms with Page References
Assignment operator, 472
Input region, 468
MuPAD Library, 480
Output region, 468
Reserved symbols, 470
Text region, 468
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Problems
Problems
You can nd the answers to problems marked with an asterisk at the end of the text.
Section 11.2
1.
Use MuPAD to prove the following identities:
a. sin2 x cos2 x 1
b. sin(x y) sin x cos y cos x sin y
c. sin2x 2 sin x cos x
d. cosh2 x sinh2 x 1
2.* Two polynomials in the variable x are represented by the coef cient
vectors p1 = [6,2,7,-3] and p2 = [10,-5,8].
a. Use MuPAD to nd the product of these two polynomials; express the
product in its simplest form.
b. Use MuPAD to nd the numeric value of the product if x 2.
3.* The equation of a circle of radius r centered at x 0, y 0 is
x2 + y2 = r2
Use MuPAD functions to nd the equation of a circle of radius r centered at
the point x a, y b. Rearrange the equation into the form Ax2 Bx Cxy Dy Ey2 F and nd the expressions for the coef cients in terms
of a, b, and r.
Section 11.3
4.* The law of cosines for a triangle states that a2 b2 c2 2bc cos A,
where a is the length of the side opposite the angle A, and b and c are the
lengths of the other sides.
a. Use MuPAD to solve for b.
b. Suppose that A 60 , a 5 m, and c 2 m. Determine b.
5.
Use MuPAD to solve the polynomial equation x3 8x2 ax 10 0
for x in terms of the parameter a, and evaluate your solution for the case
a 17. Use MuPAD to check the answer.
6.* The equation for an ellipse centered at the origin of the Cartesian
coordinates (x, y) is
y2
x2
+
= 1
a2
b2
where a and b are constants that determine the shape of the ellipse.
a. In terms of the parameter b, use MuPAD to nd the points of
intersection of the two ellipses described by
x2 +
y2
b2
= 1
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CHAPTER 11 MuPAD
and
x2
+ 4y2 = 1
100
b. Evaluate the solution obtained in part a for the case b 2.
7.
The equation
r =
p
1 - cos
describes the polar coordinates of an orbit with the coordinate origin at
the sun. If 0, the orbit is circular; if 0 < < 1, the orbit is elliptical.
The planets have orbits that are nearly circular; comets have orbits that
are highly elongated with nearer to 1. It is of obvious interest to determine whether a comet’s or an asteroid’s orbit will intersect that of a
planet. For each of the following two cases, use MuPAD to determine
whether orbits A and B intersect. If they do, determine the polar coordinates of the intersection point. The units of distance are AU, where 1 AU
is the mean distance of the Earth from the sun.
a. Orbit A: p 1, 0.01. Orbit B: p 0.1, 0.9.
b. Orbit A: p 1, 0.01. Orbit B: p 1.1, 0.5.
Section 11.4
8.
Show that R1(a)R(a) I, where I is the identity matrix and R(a) is the
rotation matrix. This equation shows that the inverse coordinate transformation returns you to the original coordinate system.
9.
Show that R1(a) R(a). This equation shows that a rotation through a
negative angle is equivalent to an inverse transformation.
10.* Find the characteristic polynomial and roots of the following matrix:
A = c
-6
3k
2
d
-7
11.* Use the matrix inverse and the matrix division method to solve the
following set for x and y in terms of c:
4cx + 5y = 43
3x - 4y = - 22
12.
The currents i1, i2, and i3 in the circuit shown in Figure P12 are described
by the following equation set if all the resistances are equal to R.
2R
-R
J
0
-R
3R
R
0 i1
␷1
- R i2 = 0
KJ K
J K
- 2R i3
␷2
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Problems
Here ␷1 and ␷2 are applied voltages; the other two currents can be found
from i4 i1 i2 and i5 i2 i3.
a. Use both the matrix inverse method and the matrix division method to
solve for the currents in terms of the resistance R and the voltages ␷1
and ␷2.
b. Find the numerical values for the currents if R 1000 , ␷1 100 V,
and ␷2 25 V.
R1
i1
i2
R2
i3
i4
R3
i5
+
+
v1
R4
R5
v2
–
–
Figure P12
13.
The equations for the armature-controlled dc motor shown in Figure P13
follow. The motor’s current is i, and its rotational velocity is .
di
= - Ri - Ke + ␷(t)
dt
d
= KTi - c
I
dt
L
where L, R, and I are the motor’s inductance, resistance, and inertia; KT
and Ke are the torque constant and back-emf constant; c is a viscous
damping constant; and ␷(t) is the applied voltage.
a. Find the characteristic polynomial.
R
L
cω
i
+
v
Ke ω
I
–
T = KT i
Figure P13
ω
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CHAPTER 11 MuPAD
14.
b. Use the values R 0.8 , L 0.003 H, KT 0.05 N m/A, Ke 0.05
V s/rad, and I 8 105 kg m2. The damping constant c is often
dif cult to determine with accuracy . For these values nd the expressions for the two characteristic roots in terms of c.
c. Using the parameter values in part b, determine the roots for the
following values of c (in N m s): c 0, c 0.01, c 0.1, and c 0.2. For each case, use the roots to estimate how long the motor’s
speed will take to become constant; also discuss whether the speed will
oscillate before it becomes constant.
Solve the following recurrence relation for the given initial conditions.
y(n 2) 0.3y(n 1) 0.02y(n) 10
15.
y(0) 2
y(1) 0
Solve the following optimization problem. Minimize
J x 3y 2z
subject to the constraints
3x + 3y - 3z … 20
6x - 4y - 3z = 20
7x + 4y + 11z … 50
Section 11.5
16.
Use MuPAD to nd all the values of x where the graph of y 3x 2x has
a horizontal tangent line.
17.* Use MuPAD to determine all the local minima and local maxima and all
the in ection points where dy兾dx 0 of the following function:
3
2
y = x4 - 16
3 x + 8x - 4
18.
The surface area of a sphere of radius r is S 4r2. Its volume is
V 4r3兾3.
a. Use MuPAD to nd the expression for dS兾dV.
b. A spherical balloon expands as air is pumped into it. What is the rate
of increase in the balloon’s surface area with volume when its volume
is 30 in.3?
19.
Use MuPAD to nd the point on the line y 2 x兾3 that is closest to the
point x 3, y 1.
20.
A particular circle is centered at the origin and has a radius of 5. Use
MuPAD to nd the equation of the line that is tangent to the circle at the
point x 3, y 4.
Ship A is traveling north at 6 mi/hr, and ship B is traveling west at 12 mi/hr.
When ship A was dead ahead of ship B, it was 6 mi away. Use MuPAD to
determine how close the ships come to each other.
21.
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Problems
22.
Suppose you have a wire of length L. You cut a length x to make a square
and use the remaining length L x to make a circle. Use MuPAD to nd
the length x that maximizes the sum of the areas enclosed by the square
and the circle.
23.* A certain spherical street lamp emits light in all directions. It is mounted
on a pole of height h (see Figure P23). The brightness B at point P on the
sidewalk is directly proportional to sin and inversely proportional to the
square of the distance d from the light to the point. Thus
c
B = 2 sin
d
where c is a constant. Use MuPAD to determine how high h should be to
maximize the brightness at point P, which is 30 ft from the base of the pole.
Light
d
h
θ
P
30 ft
Figure P23
24.* A certain object has a mass m 100 kg and is acted on by a force f (t) 500[2 et sin(5t)] N. The mass is at rest at t 0. Use MuPAD to
compute the object’s velocity ␷ at t 5 s. The equation of motion is
#
m␷ = f(t).
25. A rocket’s mass decreases as it burns fuel. The equation of motion for a
rocket in vertical ight can be obtained from Newton’ s law and is
m(t)
d␷
= T - m(t)g
dt
where T is the rocket’s thrust and its mass as a function of time is given by
m(t) m0(1 rt/b). The rocket’s initial mass is m0, the burn time is b, and
r is the fraction of the total mass accounted for by the fuel. Use the values
T 48 000 N, m0 2200 kg, r 0.8, g 9.81 m/s2, and b 40 s.
a. Use MuPAD to compute the rocket’s velocity as a function of time for
t b.
b. Use MuPAD to compute the rocket’s velocity at burnout.
26. The equation for the voltage ␷(t) across a capacitor as a function of time is
t
␷(t) =
1
c
i(t) dt + Q0 d
C L0
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CHAPTER 11 MuPAD
where i(t) is the applied current and Q0 is the initial charge. Suppose that
C 106 F and that Q0 0. If the applied current is i(t) [0.01 0.3e5t sin(25t)]103 A, use MuPAD to obtain the voltage ␷(t).
27.
The power P dissipated as heat in a resistor R as a function of the current
i(t) passing through it is P i2R. The energy E(t) lost as a function of
time is the time integral of the power. Thus
t
E(t) =
L0
t
P(t) dt = R
L0
i2(t) dt
If the current is measured in amperes, the power is in watts and the energy
is in joules (1 W 1 J/s). Suppose that a current i(t) 0.2[1 sin(0.2t)] A is
applied to the resistor.
a. Determine the energy E(t) dissipated as a function of time.
b. Determine the energy dissipated in 1 min if R 1000 .
28.
The RLC circuit shown in Figure P28 can be used as a narrowband lter . If
the input voltage ␷i(t) consists of a sum of sinusoidally varying voltages
with different frequencies, the narrowband lter will allow to pass only
those voltages whose frequencies lie within a narrow range. The magni cation ratio M of a circuit is the ratio of the amplitude of the output voltage
␷o(t) to the amplitude of the input voltage ␷i(t). It is a function of the radian
frequency of the input voltage. Formulas for M are derived in elementary
electrical circuits courses. For this particular circuit, M is given by
M =
RC
2(1 - LC2)2 + (RC)2
C
L
+
vi
R
vo
–
Figure P28
The frequency at which M is a maximum is the frequency of the desired
carrier signal. Determine this frequency as a function of R, C, and L.
29.
The shape of a cable hanging with no load other than its own weight is a
catenary curve. A particular bridge cable is described by the catenary
y(x) 10 cosh[(x 20)兾10] for 0 x 50, where x and y are the horizontal and vertical coordinates measured in feet. (See Figure P29.) It is
desired to hang plastic sheeting from the cable to protect passersby while
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Problems
the bridge is being repainted. Use MuPAD to determine how many square
feet of sheeting are required. Assume that the bottom edge of the sheeting
is located along the x axis at y 0.
y
Cable
Plastic Sheet
x
Bridge Deck
Figure P29
30.
The shape of a cable hanging with no load other than its own weight is a
catenary curve. A particular bridge cable is described by the catenary
y(x) 10 cosh[(x 20)兾10] for 0 x 50, where x and y are the horizontal and vertical coordinates measured in feet.
The length L of a curve described by y(x) for a x b can be found
from the following integral:
L
冕
a
b A
1
dy 2
冢dx冣 dx
Determine the length of the cable.
31.
Use the rst ve nonzero terms in the Taylor series for eix, sin x, and cos x
about x 0 to demonstrate the validity of Euler’s formula eix cos x i sin x.
32.
Find the Taylor series for ex sin x about x 0 in two ways:
a. By multiplying the Taylor series for ex and that for sin x.
b. By using the taylor function directly on e x sin x.
33.
Integrals that cannot be evaluated in closed form sometimes can be evaluated approximately by using a series representation for the integrand. For
example, the following integral is used for some probability calculations
(see Section 7.2):
1
I =
2
e-x dx
L0
2
a. Obtain the Taylor series for ex about x 0 and integrate the rst six
nonzero terms in the series to nd I. Use the seventh term to estimate
the error.
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CHAPTER 11 MuPAD
b. Compare your answer with that obtained with the MuPAD erf(t) function, de ned as
t
2
2
e-t dt
erf(t) =
1 L0
34.* Use MuPAD to compute the following limits.
x2 - 1
a. lim 2
x:1 x - x
x2 - 4
b. lim 2
x : -2 x + 4
x4 + 2x2
c. lim 3
x:0 x + x
35.
Use MuPAD to compute the following limits.
a. lim xx
x:0 +
b. lim (cos x)1>tan x
x:0 +
2
-1>x
1
c. lim a
b
x:0 +
1 - x
sin x2
d. lim
x : 0 - x3
x2 - 25
e. lim 2
x : 5 - x - 10x + 25
x2 - 1
f. lim
x : 1 + sin[(x - 1)2]
36.
37.
Use MuPAD to compute the following limits.
x + 1
a. lim
q
x
x:
3x3 - 2x
b. lim
x : - q 2x3 + 3
Find the expression for the sum of the geometric series
n-1
k
ar
k=0
for r
38.
1.
A particular rubber ball rebounds to one-half its original height when
dropped on a oor .
a. If the ball is initially dropped from a height h and is allowed to continue to bounce, nd the expression for the total distance traveled by
the ball after the ball hits the oor for the nth time.
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Problems
b. If it is initially dropped from a height of 10 ft, how far will the ball
have traveled after it hits the oor for the eighth time?
Section 11.6
39.
The equation for the voltage y across the capacitor of an RC circuit is
RC
dy
+ y = ␷(t)
dt
where ␷(t) is the applied voltage. Suppose that RC 0.2 s and that the
capacitor voltage is initially 2 V. If the applied voltage goes from 0 to 10 V
at t 0, use MuPAD to determine the voltage y(t).
40.
The following equation describes the temperature T(t) of a certain object
immersed in a liquid bath of temperature Tb(t):
10
dT
+ T = Tb
dt
Suppose the object’s temperature is initially T(0) 70 F and the bath
temperature is 170 F. Use MuPAD to answer the following questions:
a. Determine T(t).
b. How long will it take for the object’s temperature T to reach 168 F?
41.* This equation describes the motion of a mass connected to a spring with
viscous friction on the surface
$
#
my + cy + ky = f(t)
42.
where f(t) is an applied force. The position and velocity of the mass at
t 0 are denoted by x0 and ␷0. Use MuPAD to answer the following
questions.
a. What is the free response in terms of x0 and ␷0 if m 3, c 18, and
k 102?
b. What is the free response in terms of x0 and ␷0 if m 3, c 39, and
k 120?
The equation for the voltage y across the capacitor of an RC circuit is
RC
dy
+ y = ␷(t)
dt
where ␷(t) is the applied voltage. Suppose that RC 0.2 s and that the capacitor voltage is initially 2 V. If the applied voltage is ␷(t) 10[2 et sin(5t)],
use MuPAD to obtain the voltage y(t).
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CHAPTER 11 MuPAD
43.
The following equation describes a certain dilution process, where y(t) is
the concentration of salt in a tank of freshwater to which salt brine is
being added:
dy
2
+
y = 4
dt
10 + 2t
44.
Suppose that y(0) 0. Use MuPAD to obtain y(t).
This equation describes the motion of a certain mass connected to a
spring with viscous friction on the surface
$
#
3y + 18y + 102y = f(t)
45.
where f(t) is an applied force. Suppose that f (t) 0 for t 0 and f(t) 10
for t 0.
#
a. Use MuPAD to obtain y(t) if y(0) = y(0) = 0.
#
b. Use MuPAD to obtain y(t) if y(0) 0 and y(0) = 10.
This equation describes the motion of a certain mass connected to a
spring with viscous friction on the surface
$
#
3y + 39y + 120y = f(t)
where f(t) is an applied force. Suppose that f(t) 0 for t 0 and f(t) 10
for t 0.
#
a. Use MuPAD to obtain y(t) if y(0) = y(0) = 0.
#
b. Use MuPAD to obtain y(t) if y(0) 0 and y(0) = 10.
46.
The equations for an armature-controlled dc motor follow. The motor’s
current is i and its rotational velocity is .
di
= - Ri - Ke + ␷(t)
dt
d
= KTi - c
I
dt
L
where L, R, and I are the motor’s inductance, resistance, and inertia; KT and
Ke are the torque constant and back-emf constant; c is a viscous damping
constant; and ␷(t) is the applied voltage.
Use the values R 0.8 , L 0.003 H, KT 0.05 Nm/A, Ke 0.05
V s/rad, c 0, and I 8 105 kg · m2.
Suppose the applied voltage is 20 V. Use MuPAD to obtain the motor’s
speed and current versus time for zero initial conditions. Choose a nal time
large enough to show the motor’s speed becoming constant.
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Problems
Section 11.7
47.
48.
49.
The RLC circuit described in Problem 28 and shown in Figure P28 has
the following differential equation model:
$
#
#
LC␷ o + RC␷o + ␷o = RC␷i(t)
Use the Laplace transform method to solve for the unit-step response of
␷0(t) for zero initial conditions, where C 105 F and L 5 103 H.
For the rst case (a broadband lter), R 1000 . For the second case
(a narrowband lter), R 10 . Compare the step responses of the two
cases.
The differential equation model for a certain speed control system for a
vehicle is
#
#
$
␷ + (1 + Kp)␷ + KI␷ = Kp␷d + KI␷d
where the actual speed is ␷, the desired speed is ␷d(t), and Kp and KI are
constants called the control gains. Use the Laplace transform method to
nd the unit-step response [that is, ␷d (t) is a unit-step function]. Use zero
initial conditions. Compare the response for three cases.
a. Kp 9, KI 50
b. Kp 9, KI 25
c. Kp 54, KI 250
The differential equation model for a certain position control system for a
metal cutting tool is
d3x
d2x
dx
+ (6 + KI)x
+
(6
+
K
)
+ (11 + Kp)
D
3
2
dt
dt
dt
d2xd
dxd
= KD 2 + Kp
+ KI xd
dt
dt
where the actual tool position is x; the desired position is xd (t); and Kp, KI,
and KD are constants called the control gains. Use the Laplace transform
method to nd the unit-step response [that is, xd (t) is a unit-step function].
Use zero initial conditions. Compare the response for three cases.
a. Kp 30, KI KD 0
b. Kp 27, KI 17.18, KD 0
c. Kp 36, KI 38.1, KD 8.52
50.* The differential equation model for the motor torque m(t) required for a
certain speed control system is
$
#
#
4m + 4Km + K2m = K2␷d
where the desired speed is ␷d(t) and K is a constant called the control gain.
a. Use the Laplace transform method to nd the unit-step response
[that is, ␷d(t) is a unit-step function]. Use zero initial conditions.
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CHAPTER 11 MuPAD
b. Use symbolic manipulation in MuPAD to nd the value of the peak
torque in terms of the gain K.
Section 11.8
51.
Solve Legendre’s equation for the given initial conditions.
(1 x2)y 2xy 6y 0
52.
y(0) 0
Evaluate the solution to Airy’s equation at x 2 for the given initial
conditions.
y xy 0
53.
y(0) 5
y(0) 0
y(0) 1
Obtain a series solution to the following Legendre equation for the given
initial conditions.
(1 x2) y 2xy 6y 0
y(0) a
y(0) b
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A P P E N D I X
A
Guide to Commands
and Functions in
This Text
Operators and special characters
Item
Description
Pages
+
*
.*
^
.^
\
/
.\
./
:
Plus; addition operator.
Minus; subtraction operator.
Scalar and matrix multiplication operator.
Array multiplication operator.
Scalar and matrix exponentiation operator.
Array exponentiation operator.
Left division operator.
Right division operator.
Array left division operator.
Array right division operator.
Colon; generates regularly spaced elements and represents
an entire row or column.
Parentheses; enclose function arguments and array indices;
override precedence.
Brackets; enclose array elements.
Braces; enclose cell elements.
Ellipsis; line-continuation operator.
Comma; separates statements, and elements in a row of an array.
Semicolon; separates columns in an array, and suppresses display.
Percent sign; designates a comment, and speci es formatting.
Quote sign and transpose operator.
Nonconjugated transpose operator.
Assignment (replacement) operator.
Creates a function handle.
8
8
8
66
8
66
8, 66
8, 66
66
66
()
[]
{}
...
,
;
%
‘
.’
=
@
12, 57
9, 117
19, 55
91
12
12
12, 55
27, 549
55, 57
57
10
124
527
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Appendix A
Logical and relational operators
Item
Description
Pages
==
~=
<
<=
>
>=
&
&&
|
||
~
Relational operator: equal to.
Relational operator: not equal to.
Relational operator: less than.
Relational operator: less than or equal to.
Relational operator: greater than.
Relational operator: greater than or equal to.
Logical operator: AND.
Short-circuit AND.
Logical operator: OR.
Short-circuit OR.
Logical operator: NOT.
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155
155
155
155
155
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158
158
158
158
Special variables and constants
Item
Description
Pages
ans
eps
i,j
Inf
NaN
pi
Most recent answer.
Accuracy of oating-point precision.
The imaginary unit 1- 1.
In nity .
Unde ned numerical result (not a number).
The number .
14
14
14
14
14
14
Commands for managing a session
Item
Description
Pages
clc
clear
doc
exist
global
help
helpwin
lookfor
quit
who
whos
Clears Command window.
Removes variables from memory.
Displays documentation.
Checks for existence of le or variable.
Declares variables to be global.
Displays Help text in the Command window.
Displays Help text in the Help browser.
Searches Help entries for a keyword.
Stops MATLAB.
Lists current variables.
Lists current variables (long display).
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12
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12
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38
38
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12
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System and le commands
Item
Description
Pages
cd
date
dir
load
path
pwd
save
type
what
wk1read
xlsread
Changes current directory.
Displays current date.
Lists all les in current directory .
Loads workspace variables from a le.
Displays search path.
Displays current directory.
Saves workspace variables in a le.
Displays contents of a le.
Lists all MATLAB les.
Reads .wk1 spreadsheet le.
Reads .xls spreadsheet le.
23
122
23
21
23
23
21
38
23
138
135
Input/output commands
Item
Description
Pages
disp
format
fprintf
input
menu
;
Displays contents of an array or string.
Controls screen display format.
Performs formatted writes to screen or le.
Displays prompts and waits for input.
Displays a menu of choices.
Suppresses screen printing.
31
14, 31
549
31, 171
31
12
Numeric display formats
Item
Description
Pages
format short
format long
format short e
format long e
format bank
format +
format rat
format compact
format loose
Four decimal digits (default).
16 decimal digits.
Five digits plus exponent.
16 digits plus exponent.
Two decimal digits.
Positive, negative, or zero.
Rational approximation.
Suppresses some line feeds.
Resets to less compact display mode.
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14, 31
14, 31
14, 31
14, 31
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14, 31
14, 31
14, 31
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Array functions
Item
Description
Pages
cat
nd
length
linspace
logspace
max
min
size
sort
sum
Concatenates arrays.
Finds indices of nonzero elements.
Computes number of elements.
Creates regularly spaced vector.
Creates logarithmically spaced vector.
Returns largest element.
Returns smallest element.
Computes array size.
Sorts each column.
Sums each column.
64
60, 161
19
60
60
60
60
60
60, 296
60
Item
Description
Pages
eye
ones
zeros
Creates an identity matrix.
Creates an array of 1s.
Creates an array of 0s.
83
83
83
Special matrices
Matrix functions for solving linear equations
Item
Description
Pages
det
inv
pinv
rank
rref
Computes determinant of an array.
Computes inverse of a matrix.
Computes pseudoinverse of a matrix.
Computes rank of a matrix.
Computes reduced row echelon form.
333
333
342
335
345
Exponential and logarithmic functions
Item
Description
Pages
exp(x)
log(x)
log10(x)
sqrt(x)
Exponential; e x.
Natural logarithm; In x.
Common (base 10) logarithm; log x log10 x.
Square root; 1x.
21, 114
114
114
114
Item
Description
Pages
abs(x)
angle(x)
conj(x)
imag(x)
real(x)
Absolute value; ⱍxⱍ
Angle of a complex number x.
Complex conjugate of x.
Imaginary part of a complex number x.
Real part of a complex number x.
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114
114
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114
Complex functions
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Numeric functions
Item
Description
Pages
ceil
fix
floor
round
sign
Rounds to the nearest integer toward q
Rounds to the nearest integer toward zero.
Rounds to the nearest integer toward q
Rounds toward the nearest integer.
Signum function.
114
114
114
114
114
Item
Description
Pages
acos(x)
acot(x)
acsc(x)
asec(x)
asin(x)
atan(x)
atan2(y,x)
cos(x)
cot(x)
csc(x)
sec(x)
sin(x)
tan(x)
Inverse cosine; arccos x cos1x.
Inverse cotangent; arccot x cot1x.
Inverse cosecant; arccsc x csc1x.
Inverse secant; arcsec x sec1x.
Inverse sine; arcsin x sin1x.
Inverse tangent; arctan x tan1x.
Four-quadrant inverse tangent.
Cosine; cos x.
Cotangent; cot x.
Cosecant; csc x.
Secant; sec x.
Sine; sin x.
Tangent; tan x.
21, 118
118
118
118
21, 118
21, 118
118
21, 118
118
118
118
118
118
Item
Description
Pages
acosh(x)
acoth(x)
acsch(x)
asech(x)
asinh(x)
atanh(x)
cosh(x)
coth(x)
csch(x)
sech(x)
sinh(x)
tanh(x)
Inverse hyperbolic cosine; cosh1x.
Inverse hyperbolic cotangent; coth1x.
Inverse hyperbolic cosecant; csch1x.
Inverse hyperbolic secant; sech1x.
Inverse hyperbolic sine; sinh1x.
Inverse hyperbolic tangent; tanh1x.
Hyperbolic cosine; cosh x.
Hyperbolic cotangent; cosh x /sinh x.
Hyperbolic cosecant; 1sinh x.
Hyperbolic secant; 1cosh x.
Hyperbolic sine; sinh x.
Hyperbolic tangent; sinh x / cosh x.
119
119
119
119
119
119
119
119
119
119
119
119
Item
Description
Pages
conv
deconv
eig
poly
poly t
polyval
roots
Computes product of two polynomials.
Computes ratio of polynomials.
Computes the eigenvalues of a matrix.
Computes polynomial from roots.
Fits a polynomial to data.
Evaluates polynomial.
Computes polynomial roots.
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86
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86
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225
86
Trigonometric functions
Hyperbolic functions
Polynomial functions
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Logical functions
Item
Description
Pages
any
all
nd
nite
ischar
isinf
isempty
isnan
isreal
isnumeric
logical
xor
True if any elements are nonzero.
True if all elements are nonzero.
Finds indices of nonzero elements.
True if elements are nite.
True if elements are a character array.
True if elements are in nite.
True if matrix is empty.
True if elements are unde ned.
True if all elements are real.
True if elements have numeric values.
Converts a numeric array to a logical array.
Exclusive OR.
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161
161
161
161
161
161
161
161
161
161
161
Miscellaneous mathematical functions
Item
Description
Pages
cross
dot
function
nargin
nargout
Computes cross products.
Computes dot products.
Creates a user-de ned function.
Number of function input arguments.
Number of function output arguments.
85
85
119
170
170
Item
Description
Pages
cell
eldnames
is eld
isstruct
rm eld
struct
Creates cell array.
Returns eld names in a structure array .
Identi es a structure array eld.
Identi es a structure array .
Removes a eld from a structure array .
Creates a structure array.
90
94
94
94
94
93
Item
Description
Pages
axis
cla
fplot
ginput
grid
plot
print
title
xlabel
ylabel
Sets axis limits and other axis properties.
Clears the axes.
Intelligent plotting of functions.
Reads coordinates of the cursor position.
Displays grid lines.
Generates xy plot.
Prints plot or saves plot to a le.
Puts text at top of plot.
Adds text label to x axis.
Adds text label to y axis.
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225
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225
225
225
Cell and structure functions
Basic xy plotting commands
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Plot enhancement commands
Item
Description
Pages
colormap
gtext
hold
legend
subplot
text
Sets the color map of the current gure.
Enables label placement by mouse.
Freezes current plot.
Places legend by mouse.
Creates plots in subwindows.
Places string in gure.
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232
232
230
232
232
Item
Description
Pages
bar
loglog
plotyy
polar
semilogx
semilogy
stairs
stem
Creates bar chart.
Creates log-log plot.
Enables plotting on left and right axes.
Creates polar plot.
Creates semilog plot (logarithmic abscissa).
Creates semilog plot (logarithmic ordinate).
Creates stairs plot.
Creates stem plot.
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235
235
235
235
235
235
235
Specialized plot functions
Three-dimensional plotting functions
Item
Description
Pages
contour
mesh
meshc
meshgrid
meshz
plot3
shading
surf
surfc
sur
view
waterfall
zlabel
Creates contour plot.
Creates three-dimensional mesh surface plot.
Same as mesh with contour plot underneath.
Creates rectangular grid.
Same as mesh with vertical lines underneath.
Creates three-dimensional plots from lines and points.
Speci es type of shading.
Creates shaded three-dimensional mesh surface plot.
Same as surf with contour plot underneath.
Same as surf with lighting.
Sets the angle of the view.
Same as mesh with mesh lines in one direction.
Adds text label to z axis.
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249
246
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Program ow control
Item
Description
Pages
break
case
continue
else
elseif
end
for
if
otherwise
switch
Terminates execution of a loop.
Provides alternate execution paths within switch structure.
Passes control to the next iteration of a for or while loop.
Delineates alternate block of statements.
Conditionally executes statements.
Terminates for, while, and if statements.
Repeats statements a speci c number of times.
Executes statements conditionally.
Provides optional control within a switch structure.
Directs program execution by comparing input with
case expressions.
Repeats statements an inde nite number of times.
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188
276
166
168
166
172
165
188
188
while
183
Optimization and root- nding functions
Item
Description
Pages
fminbnd
fminsearch
fzero
Finds the minimum of a function of one variable.
Finds the minimum of a multivariable function.
Finds the zero of a function.
128
128
128
Histogram functions
Item
Description
Pages
bar
hist
Creates a bar chart.
Aggregates the data into bins.
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300
Statistical functions
Item
Description
Pages
cumsum
erf
mean
median
std
Computes the cumulative sum across a row.
Computes the error function erf(x).
Calculates the mean.
Calculates the median.
Calculates the standard deviation.
303
305
296
296
304
Random number functions
Item
Description
Pages
rand
Generates uniformly distributed random numbers
between 0 and 1; sets and retrieves the state.
Generates normally distributed random numbers;
sets and retrieves the state.
Generates random permutation of integers.
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randn
randperm
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Polynomial functions
Item
Description
Pages
poly
poly t
polyval
roots
Computes the coef cients of a polynomial from its roots.
Fits a polynomial to data.
Evaluates a polynomial and generates error estimates.
Computes the roots of a polynomial from its coef cients.
86
273
273
86
Interpolation functions
Item
Description
Pages
interp1
Linear and cubic spline interpolation of a function of
one variable.
Linear interpolation of a function of two variables.
Cubic spline interpolation.
Computes the coef cients of cubic spline polynomials.
320
interp2
spline
unmkpp
317
320
320
Numerical integration functions
Item
Description
Pages
dblquad
polyint
quad
quadl
trapz
triplequad
Numerical integration of a double integral.
Integration of a polynomial.
Numerical integration with adaptive Simpson’s rule.
Numerical integration with Lobatto quadrature.
Numerical integration with the trapezoidal rule.
Numerical integration of a triple integral.
372
371
371
371
371
371
Numerical differentiation functions
Item
Description
Pages
del2
diff(x)
Computes the Laplacian from data.
Computes the differences between adjacent
elements in the vector x.
Computes the gradient from data.
Differentiates a polynomial, a polynomial product,
or a polynomial quotient.
382
329, 382
gradient
polyder
382
382
ODE solvers
Item
Description
Pages
ode45
ode15s
odeset
Nonstiff, medium-order solver.
Stiff, variable-order solver.
Creates integrator options structure for ODE solvers.
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395
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LTI object functions
Item
Description
Pages
ss
ssdata
tf
tfdata
Creates an LTI object in state-space form.
Extracts state-space matrices from an LTI object.
Creates an LTI object in transfer-function form.
Extracts equation coef cients from an L TI object.
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400
400
400
Item
Description
Pages
impulse
initial
lsim
Computes and plots the impulse response of an LTI object.
Computes and plots the free response of an LTI object.
Computes and plots the response of an LTI object to a
general input.
Computes and plots the step response of an LTI object.
401
401
401
LTI ODE solvers
step
401
Prede ned input functions
Item
Description
Pages
gensig
Generates a periodic sine, square, or pulse input.
408
Manipulating expressions in MuPAD
Item
Description
Pages
collect
combine
delete
expand
factor
normal
rewrite
simplify
Collects terms with the same powers.
Combines terms of the same algebraic structure.
Deletes the value of an identi er .
Expands an expression.
Factors a polynomial into irreducible polynomials.
Returns the normal form of a rational expression.
Transforms an expression in terms of a target function.
Simpli es an expression.
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477
472
475
475
476
478
476
Solution of nonlinear algebraic and differential equations in MuPAD
Item
Description
Pages
assume
ode
polyroots
solve
unassume
Speci es constraints on the solution variable.
Speci es a dif ferential equation and its boundary conditions.
Numerically solves for all the roots of a polynomial.
Solves nonlinear algebraic and differential equations.
Removes constraints imposed by assume.
472, 478
502
480
480
478
Numerical evaluation in MuPAD
Item
Description
Pages
DIGITS
oat
Speci es the number of displayed digits.
Evaluates the result as a oating-point number .
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Matrix operations in MuPAD
Item
Description
Pages
charpoly
det
eigenvalues
eigenvectors
matlinsolve
matrix
rank
Finds the characteristic polynomial of a matrix.
Finds the determinant of a matrix.
Finds the eigenvalues of a matrix.
Finds the eigenvectors of a matrix.
Solves the matrix equation Ax = b.
Creates a matrix.
Finds the rank of a matrix.
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489
491
491
491
489
489
Symbolic calculus in MuPAD
Item
Description
Pages
diff
int
limit
sum
taylor
Finds the rst derivative of an expression.
Finds the single integral of an expression.
Finds the limit of an expression.
Finds the sum of terms in an expression.
Finds the Taylor series expansion of an expression.
494
497
500
500
499
Laplace transforms in MuPAD
Item
Description
Pages
invlaplace
laplace
Finds the inverse Laplace transform of an expression.
Finds the Laplace transform of an expression.
507
506
Animation functions
Item
Description
Pages
drawnow
getframe
movie
moviein
pause
Initiates immediate plotting.
Captures current gure in a frame.
Plays back frames.
Initializes movie frame memory.
Pauses the display.
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Item
Description
Pages
sound
soundsc
wavplay
wavread
wavrecord
wavwrite
Plays a vector as sound.
Scales data and plays as sound.
Plays recorded sound.
Reads Microsoft WAVE le.
Records sound from input device.
Writes Microsoft WAVE le.
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547
547
547
548
548
Sound functions
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Animation and Sound
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B.1 Animation
Animation can be used to display the behavior of an object over time. Some of
the MATLAB demos are M- les that perform animation. After completing this
section, which has simple examples, you may study the demo les, which are
more advanced. Two methods can be used to create animations in MATLAB.
The rst method uses the movie function. The second method uses the EraseMode property.
Creating Movies in MATLAB
The getframe command captures, or takes a snapshot of, the current gure to
create a single frame for the movie. The getframe function is usually used in
a for loop to assemble an array of movie frames. The movie function plays
back the frames after they have been captured.
To create a movie, use a script le of the following form.
for k = 1:n
plotting expressions
M(k) = getframe; % Saves current figure in array M
end
movie(M)
For example, the following script le creates 20 frames of the function tet/b for
0 t 100 for each of 20 values of the parameter b from b 1 to b 20.
% Program movie1.m
% Animates the function t*exp(-t/b).
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t = 0:0.05:100;
for b = 1:20
plot(t,t.*exp(-t/b)),axis([0 100 0 10]),xlabel(‘t’);
M(:,b) = getframe;
end
The line M(:,b) = getframe; acquires and saves the current gure as
a column of the matrix M. Once this le is run, the frames can be replayed as a
movie by typing movie(M). The animation shows how the location and height
of the function peak changes as the parameter b is increased.
Rotating a 3D Surface
The following example rotates a three-dimensional surface by changing the
viewpoint. The data are created using the built-in function peaks.
% Program movie2.m
% Rotates a 3D surface.
[X,Y,Z] = peaks(50);
% Create data.
surfl(X,Y,Z)
% Plot the surface.
axis([-3 3 -3 3 -5 5])% Retain same scaling for each frame.
axis vis3d off % Set the axes to 3D and turn off tick marks,
% and so forth.
shading interp
% Use interpolated shading.
colormap(winter) % Specify a color map.
for k = 1:30 % Rotate the viewpoint and capture each frame.
view(-37.5+0.5*(k-1),30)
M(k) = getframe;
end
cla
% Clear the axes.
movie(M)
% Play the movie.
The colormap(map) function sets the current gure’ s color map to map.
Type help graph3d to see a number of color maps to choose for map. The
choice winter provides blue and green shading. The view function speci es
the 3D graph viewpoint. The syntax view(az,el) sets the angle of the view
from which an observer sees the current 3D plot, where az is the azimuth or
horizontal rotation and el is the vertical elevation (both in degrees). Azimuth
revolves about the z axis, with positive values indicating counterclockwise rotation of the viewpoint. Positive values of elevation correspond to moving above
the object; negative values move below. The choice az 37.5, el 30
is the default 3D view.
Extended Syntax of the movie Function
The function movie (M) plays the movie in array M once, where M must be an
array of movie frames (usually acquired with getframe). The function
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movie(M,n) plays the movie n times. If n is negative, each “play” is once forward and once backward. If n is a vector, the rst element is the number of times
to play the movie, and the remaining elements are a list of frames to play in the
movie. For example, if M has four frames, then n = [10 4 4 2 1] plays
the movie 10 times, and the movie consists of frame 4 followed by frame 4 again,
followed by frame 2 and nally frame 1.
The function movie (M,n,fps) plays the movie at fps frames per second. If fps is omitted, the default is 12 frames per second. Computers that cannot achieve the speci ed fps will play the movie as fast as they can. The
function movie(h,...) plays the movie in object h, where h is a handle to
a gure or an axis. Handles are discussed in Section 2.2.
The function movie(h,M,n,fps,loc) speci es the location to play the
movie, relative to the lower left corner of object h and in pixels, regardless of the
value of the object’s Units property, where loc = [x y unused unused] is
a four-element position vector, of which only the x and y coordinates are used,
but all four elements are required. The movie plays back using the width and
height in which it was recorded.
Note that for code to be compatible with versions of MATLAB prior to
Release 11 (5.3), the moviein(n)function must be used to initialize movie
frame memory for n frames. To do this, place the line M = moviein(n);
before the for loop that generates the plots.
The disadvantage of the movie function is that it might require too much
memory if many frames or complex images are stored.
Animation with the EraseMode Property
One form of extended syntax for the plot function is
plot(...,‘PropertyName’,‘PropertyValue’,...)
This form sets the plot property speci ed by PropertyName to the values
speci ed by PropertyValue for all line objects created by the plot function. One such property name is EraseMode. This property controls the technique MATLAB uses to draw and erase line objects and is useful for creating
animated sequences. The allowable values for the EraseMode property are the
following.
■
normal This is the default value for the EraseMode property. By typing
plot(...,‘EraseMode’,‘normal’)
the entire gure, including axes, labels, and titles, is erased and redrawn
using only the new set of points. In redrawing the display, a three-dimensional
analysis is performed to ensure that all objects are rendered correctly. Thus,
this mode produces the most accurate picture but is the slowest. The other
modes are faster but do not perform a complete redraw and are therefore less
accurate. This method may cause blinking between each frame because
everything is erased and redrawn. This method is therefore undesirable for
animation.
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■
■
■
none When the EraseMode property value is set to none, objects in
the existing gure are not erased, and the new plot is superimposed on the
existing gure. This mode is therefore fast because it does not remove
existing points, and it is useful for creating a “trail” on the screen.
xor When the EraseMode property value is set to xor, objects are drawn
and erased by performing an exclusive OR with the background color. This
produces a smooth animation. This mode does not destroy other graphics
objects beneath the ones being erased and does not change the color of the
objects beneath. However, the object’s color depends on the background
color.
background When the EraseMode property value is set to background,
the result is the same as with xor except that objects behind the erased
objects are destroyed. Objects are erased by drawing them in the axes’
background color or in the gure background color if the axes Color
property is set to none. This damages objects that are behind the erased
line, but lines are always properly colored.
The drawnow command causes the previous graphics command to be executed immediately. If the drawnow command were not used, MATLAB would
complete all other operations before performing any graphics operations and
would display only the last frame of the animation.
Animation speed depends of the intrinsic speed of the computer and on what
and how much is being plotted. Symbols such as o, *, or + will be plotted
slower than a line. The number of points being plotted also affects the animation
speed. The animation can be slowed by using the pause(n) function, which
pauses the program execution for n seconds.
Using Object Handles
An expression of the form
p = plot(...)
assigns the results of the plot function to the variable p, which is a gure identier called a gur e handle. This stores the gure and makes it available for future
use. Any valid variable name may be assigned to a handle. A gure handle is a
speci c type of object handle. Handles may be assigned to other types of objects.
For example, later we will create a handle with the text function.
The set function can be used with the handle to change the object properties. This function has the general format
set(object handle, ‘PropertyName’, ‘PropertyValue’, ...)
If the object is an entire gure, its handle also contains the speci cations for
line color and type, marker size, and the value of the EraseMode property. Two
of the properties of the gure specify the data to be plotted. Their property
names are XData and YData. The following example shows how to use these
properties.
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Animating a Function
Consider the function tet/b, which was used in the rst movie example. This
function can be animated as the parameter b changes with the following program.
% Program animate1.m
% Animates the function t*exp(-t/b).
t = 0:0.05:100;
b = 1;
p = plot(t,t.*exp(-t/b),‘EraseMode’,‘xor’);...
axis([0 100 0 10]),xlabel(‘t’);
for b = 2:20
set(p,‘XData’,t,‘YData’,t.*exp(-t/b)),...
axis([0 100 0 10]),xlabel(‘t’);
drawnow
pause(0.1)
end
In this program the function tet/b is rst evaluated and plotted over the range
0 t 100 for b 1, and the gure handle is assigned to the variable p. This establishes the plot format for all following operations, for example, line type and
color, labeling, and axis scaling. The function tet/b is then evaluated and plotted
over the range 0 t 100 for b 2, 3, 4, . . . in the for loop, and the previous plot is erased. Each call to set in the for loop causes the next set of points
to be plotted. The EraseMode property value speci es how to plot the existing
points on the gure (i.e., how to refresh the screen), as each new set of points is
added. You should investigate what happens if the EraseMode property value is
set to none instead of xor.
Animating Projectile Motion
This following program illustrates how user-de ned functions and subplots can
be used in animations. The following are the equations of motion for a projectile
launched with a speed s0 at an angle above the horizontal, where x and y are the
horizontal and vertical coordinates, g is the acceleration due to gravity, and t is
time.
x(t) = (s0 cos )t
gt2
y(t) = - 2 + (s0 sin )t
By setting y 0 in the second expression, we can solve for t and obtain the following expression for the maximum time the projectile is in ight tmax.
tmax =
2s0
sin
g
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The expression for y(t) may be differentiated to obtain the expression for the vertical velocity:
␷vert =
dy
= - gt + s0 sin dt
The maximum distance xmax may be computed from x(tmax), the maximum height
ymax may be computed from y(tmax 2), and the maximum vertical velocity occurs
at t 0.
The following functions are based on these expressions, where s0 is the
launch speed s0 and th is the launch angle .
function x = xcoord(t,s0,th);
% Computes projectile horizontal coordinate.
x = s0*cos(th)*t;
function y = ycoord(t,s0,th,g);
% Computes projectile vertical coordinate.
y = -g*t.^2/2+s0*sin(th)*t;
function v = vertvel(t,s0,th,g);
% Computes projectile vertical velocity.
v = -g*t+s0*sin(th);
The following program uses these functions to animate the projectile motion
in the rst subplot, while simultaneously displaying the vertical velocity in the
second subplot, for the values 45, s0 105 ft/sec, and g 32.2 ft/sec2. Note
that the values of xmax, ymax, and vmax are computed and used to set the axes
scales. The gure handles are h1 and h2.
% Program animate2.m
% Animates projectile motion.
% Uses functions xcoord, ycoord, and vertvel.
th = 45*(pi/180);
g = 32.2; s0 = 105;
%
tmax = 2*s0*sin(th)/g;
xmax = xcoord(tmax,s0,th);
ymax = ycoord(tmax/2,s0,th,g);
vmax = vertvel(0,s0,th,g);
w = linspace(0,tmax,500);
%
subplot(2,1,1)
plot(xcoord(w,s0,th),ycoord(w,s0,th,g)),hold,
h1 = plot(xcoord(w,s0,th),ycoord(w,s0,th,g),’o’,‘EraseMode’,’xor’)
axis([0 xmax 0 1.1*ymax]),xlabel(‘x’), ylabel(‘y’)
subplot(2,1,2)
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plot(xcoord(w,s0,th),vertvel(w,s0,th,g)),hold,
h2 = plot(xcoord(w,s0,th),vertvel(w,s0,th,g),’s’,’EraseMode’,’xor’);
axis([0 xmax 0 1.1*vmax]),xlabel(‘x’),...
ylabel(‘Vertical Velocity’)
for t = 0:0.01:tmax
set(h1,’XData’,xcoord(t,s0,th),’YData’,ycoord(t,s0,th,g))
set(h2,’XData’,xcoord(t,s0,th),’YData’,vertvel(t,s0,th,g))
drawnow
pause(0.005)
end
hold
You should experiment with different values of the pause function
argument.
Animation with Arrays
Thus far we have seen how the function to be animated may be evaluated in the
set function with an expression or with a function. A third method is to compute
the points to be plotted ahead of time and store them in arrays. The following
program shows how this is done, using the projectile application. The plotted
points are stored in the arrays x and y.
% Program animate3.m
% Animation of a projectile using arrays.
th = 70*(pi/180);
g = 32.2; s0=100;
tmax = 2*s0*sin(th)/g;
xmax = xcoord(tmax,s0,th);
ymax = ycoord(tmax/2,s0,th,g);
%
w = linspace(0,tmax,500);
x = xcoord(w,s0,th);y = ycoord(w,s0,th,g);
plot(x,y),hold,
h1 = plot(x,y,‘o’,‘EraseMode’,‘xor’);
axis([0 xmax 0 1.1*ymax]),xlabel(‘x’),ylabel( ‘y’)
%
kmax = length(w);
for k =1:kmax
set(h1,‘XData’,x(k),‘YData’,y(k))
drawnow
pause(0.001)
end
hold
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Displaying Elapsed Time
It may be helpful to display the elapsed time during an animation. To do this,
modify the program animate3.m as shown in the following. The new lines are
indicated in bold; the line formerly below the line h1 = plot(... has been
deleted.
% Program animate4.m
% Like animate3.m but displays elapsed time.
th = 70*(pi/180);
g = 32.2; s0 = 100;
%
tmax = 2*s0*sin(th)/g;
xmax = xcoord(tmax,s0,th);
ymax = ycoord(tmax/2,s0,th,g);
%
t = linspace(0,tmax,500);
x = xcoord(t,s0,th);y = ycoord(t,s0,th,g);
plot(x,y),hold,
h1 = plot(x,y, ‘o’,‘EraseMode’,‘xor’);
text(10,10,‘Time = ‘)
time = text(30,10,‘0’,‘EraseMode’,‘background’)
axis([0 xmax 0 1.1*ymax]),xlabel(‘x’),ylabel(‘y’)
%
kmax = length(t);
for k = 1:kmax
set(h1,‘XData’,x(k),‘YData’,y(k))
t_string = num2str(t(k));
set(time,‘String’,t_string)
drawnow
pause(0.001)
end
hold
The rst new line creates a label for the time display using the text statement,
which writes the label once. The program must not write to that location again.
The second new statement creates the handle time for the text label and creates
the string for the rst time value, which is 0. By using the background value
for EraseMode, the statement speci es that the existing display of the time
variable will be erased when the next value is displayed. Note that the numerical
value of time t(k) must be converted to a string, by using the function
num2str, before it can be displayed. In the last new line, in which the set
function uses the time handle, the property name is ‘String’, which is not a
variable but a property associated with text objects. The variable being updated
is t_string.
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Appendix B
B.2 Sound
MATLAB provides a number a functions for creating, recording, and playing
sound on the computer. This section gives a brief introduction to these functions.
A Model of Sound
Sound is the uctuation of air pressure as a function of time t. If the sound is a
pure tone, the pressure p(t) oscillates sinusoidally at a single frequency, that is,
p(t) = A sin(2 ft + )
where A is the pressure amplitude (the ‘loudness”), f is the sound frequency in
cycles per second (Hz), and is the phase shift in radians. The period of the
sound wave is P 1f.
Because sound is an analog variable (one having an in nite number of values), it must be converted to a nite set of numbers before it can be stored and
used in a digital computer. This conversion process involves sampling the sound
signal into discrete values and quantizing the numbers so that they can be represented in binary form. Quantization is an issue when you are using a microphone
and analog-to-digital converter to capture real sound, but we will not discuss it
here because we will produce only simulated sounds in software.
You use a process similar to sampling whenever you plot a function in
MATLAB. To plot the function you should evaluate it at enough points to produce
a smooth plot. So, to plot a sine wave, we should “sample” or evaluate it many
times over the period. The frequency at which we evaluate it is the sampling
frequency. So, if we use a time step of 0.1 s, our sampling frequency is 10 Hz. If
the sine wave has a period of 1 s, then we are “sampling” the function 10 times
every period. So we see that the higher the sampling frequency, the better is our
representation of the function.
Creating Sound in MATLAB
The MATLAB function sound(sound_vector,sf) plays the signal in the
vector sound_vector, created with the sampling frequency sf, on the computer’s speaker. Its use is demonstrated with the following user-de ned function,
which plays a simple tone.
function playtone(freq,sf,amplitude,duration)
% Plays a simple tone.
% freq = frequency of the tone (in Hz).
% sf = sampling frequency (in Hz).
% amplitude = sound amplitude (dimensionless).
% duration = sound duration (in seconds).
t = 0:1/sf:duration;
sound_vector = amplitude*sin(2*pi*freq*t);
sound(sound_vector,sf)
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Try this function with the following values: freq = 1000, sf = 10000,
amplitude = 1, and duration = 10. The sound function truncates or “clips” any values in sound_vector that lie outside the range 1 to
1. Try using amplitude = 0.1 and amplitude = 5 to see the effect
on the loudness of the sound.
Of course, real sound contains more than one tone. You can create a sound having two tones by adding two vectors created from sine functions having different
frequencies and amplitudes. Just make sure that they are sampled with the same
frequency and, have the same number of samples, and that their sum lies in the
range 1 to 1. You can play two different sounds in sequence by concatenating
them in a row vector, as sound([sound_vector_1, sound_vector_2],
sf). You can play two different sounds simultaneously in stereo by concatenating them in a column vector, as sound([sound_vector_1’,
sound_vector_2’], sf).
MATLAB includes some sound les. For example, load the MAT- le
chirp.mat and play the sound as follows:
>>load chirp
>>sound(y,Fs)
Note that the sound vector has been stored in the MAT- le as the array y and the
sampling frequency has been stored as the variable Fs. You can also try the le
gong.mat.
A related function is soundsc(sound_vector,sf). This function
scales the signal in sound_vector to the range 1 to 1 so that the sound is
played as loudly as possible without clipping.
Reading and Playing Sound Files
The MATLAB function wavread(‘ lename’) reads a Microsoft WAVE le
having the extension .wav. The syntax is
[sound_vector, sf, bits] = wavread(‘ lename’)
where sf is the sampling frequency used to create the le and bits is the number of bits per sample used to encode the data. To play the le, use the wavplay
function as follows:
>>wavplay(sound_vector, sf)
Most computers have WAVE files to play bells, beeps, chimes, etc., to
signal you when certain actions occur. For example, to load and play the
WAVE file chimes.wav located in C:\windows\media on some PC systems,
you type
>>[sound_vector, sf] = wavread(‘c:\windows\media\chimes.wav’);
>>wavplay(sound_vector, sf)
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You can also play this sound using the sound command, as sound(y,sf), but
the wavplay function has more capabilities than the sound function. See the
MATLAB Help for information about the extended syntax of the wavplay
function.
Recording and Writing Sound Files
You can use MATLAB to record sound and write sound data to a WAVE le. The
wavrecord function records sound from a PC-based audio input device. Its
basic syntax is
sound_vector = wavrecord(n,sf)
where n is the number of samples, sampled at the rate sf. The default value is
11,025 Hz. For example, to record 5 s of audio from channel 1 sampled at 11,025 Hz,
speak into the audio device while the following program runs.
>>sf = 11025;
>>sound_vector = wavrecord(5*sf, sf);
Play back the sound by typing wavplay(sound_vector,sf).
You can use the wavwrite function to write sound stored in the vector
sound_vector to a Microsoft WAVE le. One syntax is wavwrite
(sound_vector, sf, ‘ lename’), where the sampling frequency is
sf Hz and the data are assumed to be 16-bit data. The function clips any amplitude values outside the range 1 to 1.
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C
Formatted Output
in MATLAB
The disp and format commands provide simple ways to control the screen
output. However, some users might require greater control over the screen display. In addition, some users might want to write formatted output to a data le.
The fprintf function provides this capability. Its syntax is count =
fprintf(fid,format,A,...), which formats the data in the real part of
matrix A (and in any additional matrix arguments) under control of the speci ed
format string format, and writes the data to the le associated with le identier fid. A count of the number of bytes written is returned in the variable
count. The argument fid is an integer le identi er obtained from fopen. (It
may also be 1 for standard output—the screen—or 2 for standard error. See
fopen for more information.) Omitting fid from the argument list causes output
to appear on the screen, and is the same as writing to standard output (fid 1). The
string format speci es notation, alignment, signi cant digits, eld width, and
other aspects of output format. It can contain ordinary alphanumeric characters,
along with escape characters, conversion speci ers, and other characters, or ganized as shown in the following examples. Table C.1 summarizes the basic
syntax of fprintf. Consult MATLAB Help for more details.
Suppose the variable Speed has the value 63.2. To display its value using
three digits with one digit to the right of the decimal point, along with a message,
the session is
>>fprintf(‘The speed is: %3.1f\n’,Speed)
The speed is:
63.2
Here the “ eld width” is 3, because there are three digits in 63.2. You may want
to specify a wide enough eld to provide blank spaces or to accommodate an unexpectedly large numerical value. The % sign tells MATLAB to interpret the
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Table C.1 Display formats with the fprintf function
Syntax
Description
fprintf(‘format’,A, ...)
Displays the elements of the array A, and any additional array arguments, according to the format specied in the string ‘format’.
%[][number1.number2]C, where number1
speci es the minimum eld width, number2 species the number of digits to the right of the decimal
point, and C contains control codes and format codes.
Items in brackets are optional. [-] speci es leftjusti ed.
‘format’ structure
Control codes
Format codes
Code
Description
Code
Description
\n
\r
\b
\t
‘’
\\
Start new line.
Beginning of new line.
Backspace.
Tab.
Apostrophe.
Backslash.
%e
%E
%f
%g
Scienti c format with lowercase e.
Scienti c format with uppercase E.
Decimal format.
%e or %f, whichever is shorter.
following text as codes. The code \n tells MATLAB to start a new line after displaying the number.
The output can have more than one column, and each column can have its
own format. For example,
>>r = 2.25:20:42.25;
>>circum = 2*pi*r;
>>y = [r;circum];
>>fprintf(‘%5.2f %11.5g\n’,y)
2.25
14.137
22.25
139.8
42.25
265.46
Note that the fprintf function displays the transpose of the matrix y.
Format code can be placed within text. For example, note how the
period after the code %6.3f appears in the output at the end of the displayed text.
>>fprintf(‘The rst circumference is %6.3f.\n’,circum(1))
The rst circumference is 14.137
An apostrophe in displayed text requires two single quotes. For example:
>>fprintf(‘The second circle’‘s radius %15.3e is large.\n’,r(2))
The second circle’s radius
2.225e+001 is large.
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A minus sign in the format code causes the output to be left-justi ed within its
eld. Compare the following output with the preceding example:
>>fprintf(‘The second circle’‘s radius %-15.3e is large.\n’,r(2))
The second circle’s radius 2.225e+001
is large.
Control codes can be placed within the format string. The following example
uses the tab code (\t).
>>fprintf(‘The radii are:%4.2f \t %4.2f \t %4.2f\n',r)
The radii are:
2.25
22.25
42.25
The disp function sometimes displays more digits than necessary. We can
improve the display by using the fprintf function instead of disp. Consider
the program:
p = 8.85; A = 20/100^2;
d = 4/1000; n = [2:5];
C = ((n - 1).*p*A/d);
table (:,1) = n’;
table (:,2) = C’;
disp (table)
The disp function displays the number of decimal places speci ed by the
format command (4 is the default value).
If we replace the line disp(table)with the following three lines
E=’’;
fprintf(‘No.Plates Capacitance (F) X e12 %s\n’,E)
fprintf(‘%2.0f \t \t \t %4.2f\n’,table’)
we obtain the following display:
2
3
4
5
4.42
8.85
13.27
17.70
The empty matrix E is used because the syntax of the fprintf statement
requires that a variable be speci ed. Because the rst fprintf is needed to display the table title only, we need to fool MATLAB by supplying it with a variable
whose value will not display.
Note that the fprintf command truncates the results, instead of
rounding them. Note also that we must use the transpose operation to interchange the rows and columns of the table matrix in order to display it
properly.
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Only the real part of complex numbers will be displayed with the fprintf
command. For example,
>>z = -4+9i;
>>fprintf(‘Complex number:
Complex number:
-4.00
%2.2f \n’,z)
Instead you can display a complex number as a row vector. For example,
if w 49i,
>>w = [-4,9];
>>fprintf(‘Real part is %2.0f. Imaginary part is %2.0f. \n’,w)
Real part is -4. Imaginary part is 9.
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A P P E N D I X
D
References
[Brown, 1994] Brown, T. L.; H. E. LeMay, Jr.; and B. E. Bursten. Chemistry: The
Central Science. 6th ed. Upper Saddle River, NJ: Prentice-Hall, 1994.
[Eide, 2008] Eide, A. R.; R. D. Jenison; L. L. Northup; and S. Mickelson. Introduction
to Engineering Problem Solving. 5th ed. New York: McGraw-Hill, 2008.
[Felder, 1986] Felder, R. M., and R. W. Rousseau. Elementary Principles of Chemical
Processes. New York: John Wiley & Sons, 1986.
[Garber, 1999] Garber, N. J., and L. A. Hoel. Traf c and Highway Engineering. 2nd ed.
Paci c Grove, CA: PWS Publishing, 1999.
[Jayaraman, 1991] Jayaraman, S. Computer-Aided Problem Solving for Scientists and
Engineers. New York: McGraw-Hill, 1991.
[Kreyzig, 2009] Kreyzig, E. Advanced Engineering Mathematics. 9th ed. New York:
John Wiley & Sons, 1999.
[Kutz, 1999] Kutz, M., editor. Mechanical Engineers’ Handbook. 2nd ed. New York:
John Wiley & Sons, 1999.
[Palm, 2010] Palm, W. System Dynamics. 2nd ed. New York: McGraw-Hill, 2010.
[Rizzoni, 2007] Rizzoni, G. Principles and Applications of Electrical Engineering.
5th ed. New York: McGraw-Hill, 2007.
[Star eld, 1990] Star eld, A. M.; K. A. Smith; and A. L. Bleloch. How to Model It:
Problem Solving for the Computer Age. New York: McGraw-Hill, 1990.
553
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Answers to Selected
Problems
Chapter 1
2. (a) ⫺13.3333; (b) 0.6; (c) 15; (d) 1.0323.
8. (a) x ⫹ y ⫽ ⫺3.0000 ⫺ 2.0000i; (b) xy ⫽
⫺13.0000 ⫺ 41.0000i; (c) x兾y ⫽ ⫺1.7200
⫹ 0.0400i.
18. x ⫽ ⫺15.685 and x ⫽ 0.8425 ⫾ 3.4008i.
14.
25.
28.
31.
Chapter 2
3.
A = c
38.
0
-20
6
-10
12
0
18
10
24
20
30
d
30
7.
(a) Length is 3. Absolute values ⫽ [2 4 7];
(b) Same as (a); (c) Length is 3. Absolute
values ⫽ [5.8310 5.0000 7.2801].
11. (b) The largest elements in the rst, second, and
third layers are 10, 9, and 10, respectively.
The largest element in the entire array is 10.
12.
(a)
A + B + C = c
(b) A - B + C = c
13.
-6
23
-3
d
15
-14 7
d
- 1 19
(a) A.*B = [784, -128; 144,32];
(b) A/B = [76, -168; -12, 32];
(c) B.^3 = [2744, -64;216, -8].
39.
(a) F.*D = [1200, 275, 525, 750,
3000] J; (b) sum(F.*D) = 5750 J.
(a) A*B = [-47, -78; 39, 64];
(b) B*A = [-5, -3, 48, 22].
60 tons of copper, 67 tons of magnesium, 6 tons
of manganese, 76 tons of silicon, and 101 tons
of zinc.
M ⫽ 675 N⭈m if F is in newtons and r is in
meters.
[q,r] = deconv([14,-6,3,9], [5,7,4]), q = [2.8, -5.12], r = [0, 0,
50.04, -11.48]. The quotient is 2.8x ⫺ 5.12
with a remainder of 50.04x ⫺ 11.48.
2.0458.
Chapter 3
1. (a) 3, 3.1623, 3.6056;
(b) 1.7321i, 0.2848 ⫹ 1.7553i, 0.5503 ⫹ 1.8174i;
(c) 15 ⫹ 21i, 22 ⫹ 16i, 29 ⫹ 11i;
(d) ⫺0.4 ⫺ 0.2i, ⫺0.4667 ⫺ 0.0667i,
⫺0.5333 ⫹ 0.0667i.
2. (a) ⱍxyⱍ = 105, ∠xy = - 2.6 rad;
(b) ⱍx>yⱍ = 0.84, ∠ x>y = - 1.67 rad.
3. (a) 1.01 rad (58⬚); (b) 2.13 rad (122⬚);
(c) ⫺1.01 rad (⫺58⬚); (d ) ⫺2.13 rad (⫺122⬚).
7. F1 ⫽ 197.5217 N.
10. 2.7324 sec while ascending; 7.4612 sec while
descending.
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Answers to Selected Problems
Chapter 4
4. (a) z = 1; (b) z = 0; (c) z = 1; (d ) z = 1.
5. (a) z = 0; (b) z = 1; (c) z = 0; (d ) z = 4;
(e) z = 1; ( f ) z = 5; (g) z = 1; (h) z = 0.
6. (a) z = [0, 1, 0, 1, 1];
(b) z = [0, 0, 0, 1, 1];
(c) z = [0, 0, 0, 1, 0];
(d ) z = [1, 1, 1, 0, 1].
11. (a) z = [1, 1, 1, 0, 0, 0];
(b) z = [1, 0, 0, 1, 1, 1];
(c) z = [1, 1, 0, 1, 1, 1];
(d ) z = [0, 1, 0, 0, 0, 0].
13. (a) $7300; (b) $5600; (c) 1200 shares;
(d ) $15,800.
28. Best location: x ⫽ 9, y ⫽ 16. Minimum cost:
$294.51. There is only one solution.
34. After 33 years, the amount will be $1,041,800.
36. W ⫽ 300 and T ⫽ [428.5714, 471.4286,
266.6667, 233.3333, 200, 100]
48. Weekly inventory for cases (a) and (b):
Week
Inventory (a)
Inventory (b)
1
50
30
2
50
25
3
45
20
4
40
20
5
30
10
Week
Inventory (a)
Inventory (b)
6
30
10
7
30
5
8
25
0
9
20
0
10
10
(⬍0)
Chapter 5
1. Production is pro table for Q ⱖ 108 gal/yr. The
pro t increases linearly with Q, so there is no
upper limit on the pro t.
3. x ⫽ ⫺0.4795, 1.1346, and 3.8318.
5. 37.622 m above the left-hand point, and
100.6766 m above the right-hand point.
10. 0.54 rad (31⬚).
14. The steady-state value of y is y ⫽ 1. y ⫽ 0.98
at t ⫽ 4兾b.
17. (a) The ball will rise 1.68 m and will travel
9.58 m horizontally before striking the ground
after 1.17 s.
Chapter 6
2. (a) y ⫽ 53.5x ⫺ 1354.5;
(b) y ⫽ 3582.1x⫺0.9764;
(c) y ⫽ 2.0622 ⫻ 105(10)⫺0.0067x
555
(a) b ⫽ 1.2603 ⫻ 10⫺4; (b) 836 years.
(c) Between 760 and 928 years ago.
8. If unconstrained to pass through the origin,
f ⫽ 0.3998x ⫺ 0.0294. If constrained to pass
through the origin, f ⫽ 0.3953x.
10. d ⫽ 0.0509␷ 2 ⫹ 1.1054␷ ⫹ 2.3571; J ⫽ 10.1786;
S ⫽ 57,550; r2 ⫽ 0.9998.
11. y ⫽ 40 ⫹ 9.6x1 ⫺ 6.75x2. Maximum percent
error is 7.125 percent.
4.
Chapter 7
7. (a) 96%; (b) 68%.
11. (a) Mean pallet weight is 3000 lb. Standard
deviation is 10.95 lb; (b) 8.55 percent.
18. Mean yearly pro t is $64,609. Minimum
expected pro t is $51,340. Maximum expected
pro t is $79,440. Standard deviation of yearly
pro t is $5967.
22. The estimated temperatures at 5 P.M. and 9 P.M.
are 22.5⬚ and 16.5⬚.
Chapter 8
2. (a) C ⴝ B⫺1(A⫺1B ⴚ A).
(b) C = [-0.8536, -1.6058; 1.5357,
1.3372].
5. (a) x ⫽ 3c, y ⫽ ⫺2c, z ⫽ c; (b) The plot consists
of three straight lines that intersect at (0,0).
8. T1 ⫽ 19.7596⬚C, T2 ⫽ ⫺7.0214⬚C,
T3 ⫽⫺ 9.7462⬚C. Heat loss in watts is 66.785.
11. In nite number of solutions: x ⫽ ⫺1.3846z ⫹
4.9231, y ⫽ 0.0769z ⫺ 1.3846.
14. Unique solution: x ⫽ 8 and y ⫽ 2.
15. Least-squares solution; x ⫽ 6.0928 and
y ⫽ 2.2577.
Chapter 9
1. 23 690 m.
7. 13.65 ft.
10. 1363 m/s.
25. 150 m/s.
Chapter 11
2. (a) 60x5 ⫺ 10x4 ⫹ 108x3 ⫺ 49x2 ⫹ 71x ⫺ 24;
(b) 2546.
3. A ⫽ 1, B ⫽ ⫺2a, C ⫽ 0, D ⫽ ⫺2b, E ⫽ 1, and
F ⫽ r2 ⫺ a2 ⫺ b2.
pal34870_ans_554-556.qxd
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6.
Page 556
Answers to Selected Problems
556
4.
9:49 AM
(a) b = c cos A ⫾ 2a2 - c2 sin2A; (b) b ⫽ 5.6904.
2
2
(a) x = ⫾ 102(4b - 1)>(400b - 1),
y = ⫾ 299>(400b2 - 1);
10.
11.
(b) x ⫽ ⫾0.9685, y ⫽ ⫾0.4976.
s2 ⫹ 13s ⫹ 42 ⫺ 6k, s = ( -13 ⫾ 11 + 24k)>2.
x =
62
16c + 15
y =
129 + 88c
16c + 15
23.
24.
34.
41.
␪ ⫽ 0.6155 rad (35.26⬚).
49.6808 m/s.
(a) 2; (b) 0; (c) 0.
(a) (3x0>5 + ␷0>5)e-3t sin 5t + x0 e-3t cos 5t;
(b) e-5t(8x0>3 + ␷0>3) + (-5x0>3 - ␷0>3)e-8t
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Page 557
INDEX
MATLAB Symbols
⫹ addition, 8
⫺ subtraction, 8
* multiplication, 8
.* array multiplication,
66
^ exponentiation, 8
.^ array exponentiation,
66
\ left division, 8, 66
/ right division, 8, 66
.\ array left division, 66
./ array right division, 66
: colon
array addressing,
57, 58
array generation, 12,
54, 55
() parentheses
function arguments,
117
modifying
precedence, 9
{ } braces; enclose cell
elements, 91
[] brackets, 19, 55
. . . ellipsis, line
continuation, 12
, comma
column separation,
12
statement separation,
12
; semicolon
display suppression,
12
row separation, 55
% percent sign
comment designation, 27
format speci cation,
549
’ apostrophe
complex conjugate
transpose, 57
string designation,
31, 171
transpose, 55
.’ nonconjugated
transpose (dot
transpose), 57
⫽ assignment or
replacement
operator, 10
⫽⫽ equal to, 155
~⫽ not equal to, 155
⬍ less than, 155
⬍⫽ less than or equal
to, 155
⬎ greater than, 155
⬎⫽ greater than or
equal to, 155
& AND, 158
&& short-circuit AND,
158
| OR, 158
|| short-circuit OR, 158
~NOT, 158
>> MATLAB prompt,
6
@ creates a function
handle, 124
cell, 90
celldisp, 91
cellplot, 91
cla, 539
clabel, 249
clc, 12
clear, 12
colormap, 539
conj, 114
continue, 276
contour, 250
conv, 86
cos, 21, 118
cosh, 119
cot, 118
coth, 119
cross, 85
csc, 118
csch, 119
cumsum, 303
MATLAB Commands
A
abs, 114
acos, 21, 118
acosh, 119
acot, 118
acoth, 119
acsc, 118
acsch, 119
addpath, 23
all, 161
angle, 114
ans, 14
any, 161
asec, 118
asech, 119
asin, 21, 118
asinh, 119
atan, 21, 118
atan2, 118
atanh, 119
axis, 222, 225
B
bar, 235, 295, 300
break, 176
bvp4c, 409
C
case, 188
cat, 64
cd, 23
ceil, 114
D
date, 122
dblquad, 372
dde23, 409
ddesd, 409
deconv, 86
del2, 382
det, 333
deval, 409
diff, 379, 382
dir, 23
disp, 31
557
pal34870_index_557-564.qxd
558
doc, 38
dot, 85
drawnow, 541
E
eig, 397
else, 166
elseif, 168
end, 166
eps, 14
erf, 305
exist, 12
exp, 21, 114
eye, 83
F
eldnames, 94
nd, 60, 161
nite, 161
x, 114
oor, 114
fminbnd, 128
fminsearch, 128
for, 172
format, 15, 31
fplot, 223, 225
fprintf, 549
function, 119
fzero, 128
1/9/10
5:59 PM
Page 558
Index
I
i, 14
if, 165
imag, 114
impulse, 401
Inf, 14
initial, 401
inline, 130
input, 31, 171
interp1, 317, 320
interp2, 317
interpn, 317
inv, 333
ischar, 161
isempty, 161
is eld, 94
isinf, 161
isnan, 161
isnumeric, 161
isreal, 161
isstruct, 94
J
j, 14
L
gensig, 408
getframe, 538
ginput, 25
gradient, 382
grid, 25, 222, 225
global, 124
gtext, 25, 232
legend, 232
length, 19, 60
linspace, 56, 60
load, 21
log, 21, 114
log10, 21, 114
logical, 156, 161
loglog, 235
logspace, 56, 60
lookfor, 38
lsim, 401
H
M
help, 38
helpwin, 38
hist, 296, 300
hold, 231, 232
max, 60
mean, 296
median, 296
menu, 31
G
mesh, 250
meshc, 250
meshgrid, 250
meshz, 250
min, 60
mode, 296
movie, 538
moviein, 538
mupadwelcome, 466
N
NaN, 14
nargin, 170
nargout, 170
norm, 60
O
ode15i, 409
ode15s, 385, 395
ode45, 385, 395
odephase, 409
odeplot, 409
odeprint, 409
odeset, 394, 395
ones, 83
otherwise, 188
P
path, 23
pathtool, 23
pause, 541
pchip, 320, 329
pdeval, 409
pi, 14
pinv, 342
plot, 25, 225, 232
plotyy, 235
plot3, 246
polar, 235
poly, 86
polyder, 382
poly t, 266, 273
polyint, 371
polyval, 86, 87,
225, 273
print, 221, 225
pwd, 23
Q
quad, 371
quadl, 371
quit, 12
quiver, 382
R
rand, 307, 309
randn, 309, 310
randperm, 309
rank, 335
real, 114
rm eld, 94
rmpath, 23
roots, 20, 86
round, 114
rref, 345
S
save, 21
sec, 118
sech, 119
semilogx, 235
semilogy, 235
shading, 539
sign, 114
simulink, 431
sin, 21, 118
sinh, 119
size, 60
sort, 60, 296
sound, 546
soundsc, 547
spline, 318, 320
sqrt, 21, 114
ss, 399, 400
ssdata, 399, 400
stairs, 235
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Page 559
Index
std, 304
stem, 235
step, 401
struct, 94
subplot, 232
sum, 60
surf, 250
surf1, 539
surfc, 250
switch, 188
text, 232
tf, 399, 400
tfdata, 400
title, 25, 225
trapz, 371
triplequad, 371
type, 38
T
V
tan, 21, 118
tanh, 119
var, 304
view, 539
U
unmkp, 319, 320
W
X
waterfall, 250
wavplay, 547
wavread, 547
wavrecord, 548
wavwrite, 548
what, 23
which, 23
while, 183
who, 12
whos, 12
wk1read, 138
xlabel, 25, 225
xlsread, 135
xor, 159, 161
559
Y
ylabel, 25, 225
Z
zeros, 83
zlabel, 248
Simulink Blocks
C
I
S
T
Clock, 425
Constant, 432
Integrator, 420
Saturation, 450
Scope, 425
Signal Builder, 453
Signal Generator, 443
Sine Wave, 423
State-Space, 427
Step, 429
Subsystem, 443
Summer, 421
To Workspace, 425
Transfer Fcn, 437
Transfer Fcn (with
initial outputs),
437
Transport Delay, 448
Trigonometric
Function, 432
D
L
Look-Up Table, 454
Dead Zone, 438
Derivative, 453
M
F
MATLAB Fcn, 454
Mux, 425
Fcn, 454
G
Gain, 420
R
Rate Limiter, 450
Relay, 433
MuPAD Symbols and Commands1
Symbols
: colon, display
suppression, 470
:⫽ assignment
operator, 472
:: library reference,
480
% reference to
previous result,
471
´ derivative notation,
502
# place marker, 475
Commands
A
B
airyAi(x), 512
airyBi(x), 512
arg, 471
assume, 478
besselI, 512
besselJ, 512
besselK, 512
besselY, 512
bool, 476
Most MuPAD symbols and commands are identical to their MATLAB counterparts; for example cos and cos. Here we list commonly
used symbols and commands that have special meaning in MuPAD.
1
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Page 560
Index
C
G
M
charpoly, 489
chebyshev1, 512
collect, 478
combine, 477
conjugate, 471
gamma, 512
matlinsolve, 491
matrix, 489
maximize, 488
minimize, 489
D
DIGITS, 471
delete, 472
det, 489
diff, 494
E
E, 470
eigenvalues, 491
eigenvectors, 491
expand, 475
F
factor, 475
factorout, 476
oat , 471
H
heaviside, 508
hermite, 512
realroot, 497
rec, 487
rectform, 472
rewrite, 478
N
S
I
normal, 476
I, 470
Im, 471
int, 497
inverse, 491
invlaplace, 507
ode, 502
op, 514
series, 513
Simplify, 476
simplify, 476
solve, 480, 502
subs, 479
sum, 500
L
laguerreL, 512
laplace, 506
legendre, 512
limit, 500
log, 470
ln, 470
O
P
pdioe, 487
PI, 470
plotfunc2d, 514
polyroots, 480
T
R
unassume, 478
taylor, 499
U
rank, 489
Re, 471
Topics
A
absolute frequency,
297
absolute value, 61
algebraic equations,
see also linear
algebraic equations
polynomial diophantine equations,
487
recurrence relations,
488
solving numerically,
483
solving sets of,
483
solving symbolically, 480
algorithm, 148
animation, 538
anonymous function,
130, 132
argument, 8
array, 19
addition and subtraction, 65
addressing, 58
cell, 90
creating an, 55
division, 69
empty, 58
exponentiation, 70
functions, 60
index, 19
multidimensional, 63
multiplication, 65
operations, 65
pages, 63
powers, 70
size, 56
structure, 93
ASCII les, 21
assignment operator,
10
augmented matrix, 335
axis label, 220
axis limits, 251
B
backward differences,
378
bar plots, 296
Basic Fitting Interface,
282
bins, 296
block diagram, 420
Block Parameters
window, 424
Boolean operator, 157
boundary value
problem, 409
breakpoint, 193
C
Cauchy form, 390
cell indexing, 90
cell array, 90
cell mode, 191
central difference, 379
characteristic roots,
397
clearing variables, 12
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Page 561
Index
coef cient of
determination, 275
colors, 228
column vector, 54
command, 8
Command bar
(MuPAD), 475
Command window, 6
comment, 27
common mathematical
functions, 21, 114
complex numbers, 16
complex conjugate
transpose, 57
computer solution
steps, 41, 149
conditional statement,
164
content indexing, 90
contour plots, 248
Control System
toolbox, 398
cubic splines, 317
current directory, 16
curve t, quality of, 275
D
data les, 21
data markers, 25, 228
Data Statistics tool, 300
data symbol, 251
dead time, 448
dead zone, 437
Debug menu, 192
debugging, 29, 190
de nite integral, 370
delay differential
equation, 409
derivative, see
differentiation
Desktop, 5
determinants, 335
differential equations
Cauchy form, 391
characteristic roots,
397
delay, 409
higher order, 390
nonlinear, 383
ordinary, 382
partial, 382
piecewise-linear, 430
solvers, 385
state variable form,
391
symbolic solution of,
501
differentiation,
numerical, 329, 382
partial, 409
polynomial, 382
symbolic, 494
directory, 6
dot transpose, 57
E
Edit menu, 17
Editor/Debugger, 28,
190
eigenvalue, 397, 491
element-by-element
operations, 66,
69, 70
ellipsis, 12
empty array, 58
EraseMode property,
540
error function, 305
Euclidean norm, 343
Euler method, 383
exporting data, 139
exporting gures, 225
extrapolation, 313
F
eld, 92
gure handle, 540
File menu, 17
les
ASCII, 21
command, 27
data, 21
function, 27
MAT- les, 20
M- les, 20
script, 27
spreadsheet, 135
user-de ned, 1 19
owchart, 150
for loop, 172
forced response, 383
formatting, 15
forward differences,
379
free response, 383
function argument,
117
function de nition line,
119
function discovery,
263
function le, 1 19
function handle, 124
functions
anonymous, 131
argument, 117
complex, 114
elementary mathematical, 113
exponential, 114
handle, 124
hyperbolic, 118
logarithmic, 124
minimization of,
126, 127
nested, 131
numeric, 116
overloaded, 131
primary, 131
private, 131
of random variables,
311
561
subfunction, 131
trigonometric, 117
user-de ned, 1 19
zeros of, 124
G
Gauss elimination, 335
Gaussian function, 303
global variable, 124
gradient, 380
Graphics window, 23
grid, 25, 222
H
H1 line, 30
handle, 124
help functions, 38
Help system, 33
histogram, 296
histogram functions,
300
homogeneous
equations, 335
hyperbolic functions,
119
I
identity matrix, 82
ill-conditioned
problem, 333
implied loop, 177
importing data, 172
importing spreadsheet
les, 173
Import Wizard, 173
improper integral, 370
inde nite integral,
370
initial value problem,
382
input region, 468
input/output
commands, 31
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Index
integral,
de nite, 370
double, 376
improper, 370
inde nite, 370
singularity, 370
triple, 377
integration
numerically, 371
symbolically, 497
panel, 370
trapezoidal, 370
interpolation, 313
cubic spline, 317
Hermite polynomials, 329
linear, 315, 317
2–D, 316
polynomial, 320
inverse Laplace
transform, 507
L
Laplace transform, 506
Laplacian, 382
least squares, 312
left division method, 8,
335
legend, 230
Library Browser, 421
limits, 500
line continuation, 12
line types, 228
linear algebra
characteristic polynomial, 491
eigenvalues, 397,
491
eigenvectors, 491
matrix operations,
489
linear algebraic
equations, 26, 84,
331
application of matrix
rank, 335
and augmented
matrix, 335
and Euclidean norm,
343
homogeneous, 335
ill-conditioned
system of, 333
and linearity, 336
matrix solution, 492
overdetermined
system of, 350
and reduced row
echelon form, 345
singular set of, 333
solution by left division method, 28,
335
solution by matrix
inverse, 333
solution by pseudoinverse method,
342
underdetermined
system of, 341
linear-in-parameters,
329
linear interpolation
functions, 317
linear programming,
488
local variable, 32, 119
logarithmic plots,
233
logical arrays, 157
logical functions, 161
logical operators, 158
logical variable, 123,
155
loop variable, 172
LTI ODE solvers,
401
LTI object, 399, 400
LTI Viewer, 407
M
M- les, 29
magnitude, 61
managing the work
session, 12
mask, 180
MAT- les, 20
MathWorks website,
37
matrix, 56
augmented, 345
creating a, 55
division, 83
exponentiation, 84
identity, 82
inverse, 332
multiplication,
75, 80
null, 82
operations, 73
rank, 334
special, 82
transpose, 57
unity, 82
mean, 296
median, 296
methodology
for developing a
computer solution,
42
for engineering problem solving, 38
minimization and rootnding functions,
128
modi ed Euler
method, 384
multidimensional
arrays, 81
multiple linear
regression, 328
MuPAD libraries, 480
MuPAD menus
Combine, 477
General math, 475
Rewrite, 478
Simplify, 476
Solve, 481
N
naming variables, 11
nested function, 131,
135
nested loops, 174
normal distribution, 301
normal function, 303
normally distributed
numbers, 303
null matrix, 82
numeric display
formats, 15
numerical
differentiation,
329, 382
numerical integration
functions, 371
O
ODE. See differential
equation, ordinary
operations research, 193
optimization problems,
488
order of precedence, 9,
158
output region, 468
overdetermined
system, 350
overlay plots, 24, 228
overloaded function,
131
P
pages (in
multidimensional
arrays), 81
panel, 471
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Page 563
Index
path, 22
PI controller, 450
plot, 25
axis label, 220
bar, 296
colors, 228
contour, 248
data markers, 228
editor, 241
enhancement commands, 232
hints for improving,
220
grid, 25, 222
interactive interface,
241
legend, 230
line types, 228
logarithmic, 233
overlay, 24, 228
polar, 236
requirements, 221
second y-axis, 236
specialized, 235
stairs, 230
stem, 236
subplots, 226
surface mesh, 247
text placement, 232
three-dimensional,
250
three-dimensional
line, 246
tick mark label, 220
tick marks 220
tick mark spacing,
220
title, 220, 225
plotting
complex numbers,
223
in MuPAD, 473
polynomials, 87
with smart function
plot command, 227
symbolic expressions, 593
tools, 243
xy plots, 225
polar plot, 235
polynomial, 20
addition, 86
differentiation, 380
division, 86
functions, 86
interpolation functions, 320
multiplication, 86
plotting, 87
regression, 273
roots, 20
precedence 9, 158
prede ned constants, 15
prede ned input
functions, 407
predictor-corrector
method, 384
primary function, 131
private function, 137
program documentation,
149
programming style, 30
prompt, 6
pseudocode, 151
pseudoinverse method,
342
Q
quadrature, 373
R
random number
functions, 309
random number
generator, 307
rank, 334
rate limiter, 450
rectangular integration,
370
reduced form, 390
reduced row echelon
form, 345
regression, 271
relational operators, 155
relative frequency, 287
relay, 433
replacement operator,
10
reserved symbols in
MuPAD, 470
residuals, 272, 277
right division, 8
row vector, 54
r-squared value, 275
Runge-Kutta methods,
385
S
saturation nonlinearity,
440
saving gures, 225
session, 7
scalar, 8
scalar arithmetic
operations, 8
scaled frequency
histogram, 301
scaling data, 276
script le, 27
search path, 2
session, 7
short-circuit operators,
160
simulation, 193
simulation diagrams,
420
singular matrix, 333
singularity, 370
smart recall, 13
sound, 546
special functions
(MuPAD), 512,
513
563
special matrices, 83
special variables and
constants, 14
spreadsheet les, 138
stairs plots, 235
standard deviation,
303
state of random
generator, 307
state transition
diagram, 195
state-variable form,
390, 441
statically indeterminate
problem, 343
stem plots, 235
step function, 402
step size, 384
string, 31
structure arrays, 92
structure chart, 150
structure functions,
94
structured
programming, 148
subdeterminant, 335
subfunction, 131, 134
subplots, 232
subsystems, 443
surface mesh plot,
247
sums, 500
switch structure, 188
symbolic expressions,
472
collecting, 478
combining, 477
evaluating, 476
expanding, 475
factoring, 475
manipulating, 474
normalizing, 476
rewriting, 478
simplifying, 476
substituting, 479
pal34870_index_557-564.qxd
564
system, directory, and
le commands, 23
T
tab completion, 13
Taylor series, 499
text region, 468
three-dimensional
plots, 246
contour plots, 248
line plots, 246
surface mesh plots,
247
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Page 564
Index
toolbox, 5
top-down design,
149
transfer-function form,
437
transport delay,
448
transpose, 55
trapezoidal integration,
370
trigonometric
functions, 118, 119
truth table, 159
U
underdetermined
system, 341
uniformly distributed
numbers, 307
user-de ned functions,
119
V
Variable editor, 62
variance, 303
variable, 7
vector, 70
absolute value of, 61
cross product, 85
dot product, 85
length of, 61
magnitude of, 61
multiplication, 74
W
while loop, 183
working directory. See
current directory
workspace, 11